# NCERT Solutions Class 12 Mathematics Chapter 8

Integral Calculus is related to Eudoxus' (440 B.C.) and Archimedes' (440 B.C.) exhaustion approach (300 B.C.). This method emerged from the need to solve problems estimating the areas of plane figures, surface areas, and solid body volumes. Newton began working on Calculus in 1665, which led to the formulation of concepts as we know them. NCERT solutions class 12 mathematics chapter 8 provides various hints, tactics, and intriguing facts to help students break up the monotony of studying and develop a mathematical base that will help them throughout their studies.

NCERT solutions class 12 mathematics chapter 8 - Application of Integrals is a continuation of the previous chapter, in which students learned about integration and how to solve problems using various methods. Students have learnt formulae for calculating the areas of triangles, rectangles, trapeziums, and circles, among other geometrical figures. These fundamental formulae are crucial in applying mathematics to a wide range of real-world issues. However, this method fails when estimating the areas encompassed by curves. Integral calculus is required to calculate the areas under a curve. NCERT solutions class 12 mathematics chapter 8 use integrals to get the area under curves. Curves such as circles, parabolas, and hyperbolas can be simple or complex.

### Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 8

All the topics covered in NCERT Solutions Class 12 Mathematics Chapter 8 are equally important. They go over many aspects of integration and how to calculate the area under a curve throughout multiple scenarios. Before attempting the practice questions, students must first grasp the underlying concepts. They should also be familiar with the preceding chapter and the formulas listed. The  NCERT Solutions Class 12 Mathematics Chapter 8 broadly covers two main concepts of the chapter 'Application of Integrals' and  'Integral Calculus'.

### List of NCERT Solutions Class 12 Mathematics Chapter 8 Exercises

Students learned about definite integrals as the limit of a sum in Chapter 7 and how to evaluate them using the Fundamental Theorem of Calculus. This concept is used in NCERT solutions for class 12 mathematics chapter 8 to give students an easy approach to computing the area under curves. The area beneath the curve can be thought of as a collection of several very thin vertical stripes. We may now determine the total area necessary by integrating this region between some specific limitations. NCERT Solutions for class 12 mathematics chapter 8 Application of Integrals gives students a better grasp of integration using real-life examples. We'll look at a range of examples, formulas, and theories in depth in this article to help students create a strong mathematical foundation.

Several formulas and theorems are used in this lesson. There is also a blend of current topics with those that the students learned in the previous chapter. If not approached systematically, integration can become a difficult and perplexing concept. Students must regularly review this lesson for all these reasons, particularly the formulas, to learn the material. With the help of examples and practice problems, all the tasks in NCERT Solutions Class 12 Mathematics Chapter 8 - Application of Integrals are well explained. Here are some NCERT solutions for students to practise –

mathematics Chapter 8 Ex 8.1 - 13 Questions

Class 12 Mathematics Chapter 8 Ex 8.2 - 7 Questions

Class 12 Mathematics Chapter 8 Miscellaneous Ex - 19 Questions

The area bounded by the curve of a function, the area between two curves, and the area under simple curves are all covered in NCERT solutions class 12 mathematics chapter 8.

NCERT Solutions Class 12 Mathematics Chapter 8 Formula List

NCERT solutions class 12 mathematics chapter 8 requires using formulas and a constant focus. Students must comprehend what the question requires and carry out the procedures stepwise. They should also divide the problem into smaller parts to make it easier to understand and avoid ambiguity, hence lowering the risk of errors. More significantly, children must understand the processes in the technique since they are the keys to answering a variety of questions.

The following are some formulas that are regularly used in NCERT solutions for class 12 mathematics chapter 8 –

• Area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) : a∫by dx = a∫bf(x) dx.
• The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d : c∫d x dy = c∫d φ (y) dy.
• The area of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b : b∫a[f(x) - g(x)] dx.

Class 12 NCERT Mathematics Syllabus

Term - 1

 Unit Name Chapter Name Relations and Function Relations and Functions Inverse Trigonometric Functions Algebra Matrices Determinants Calculus Continuity and Differentiability Application of Derivatives Linear Programming Linear Programming

Term - 2

 Unit Name Chapter Name Calculus Integrals Application of Integrals Differential Equations Vectors and Three-Dimensional Geometry Vector Algebra Three Dimensional Geometry Probability Probability

Experts at Extramarks create NCERT Solutions to assist students to understand ideas more quickly and correctly. NCERT Solutions provide extensive, step-by-step explanations of textbook problems. All classes can benefit from such solutions –

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NCERT Mathematics Exam Pattern

 Duration of Marks 3 hours 15 minutes Marks for Internal 20 marks Marks for Theory 80 marks Total Number of Questions 38 Questions Very short answer question 20 Questions Short answer questions 7 Questions Long Answer Questions (4 marks each) 7 Questions Long Answer Questions (6 marks each) 4 Questions

### Key Features of NCERT Solutions Class 12 Mathematics Chapter 8

To do well in CBSE and pass various competitive exams such as NEET and IIT, one must have a strong mathematical foundation. The NCERT books present you with difficult problems that improve your analytical ability and expose you to many questions that could appear in all of these exams. Our integrals class 12 solutions adhere to the most recent CBSE curriculum and are incredibly valuable to your preparation for the following reasons:

• The answers are exceptional because teachers with deep subject knowledge create them.
• You can revise all the significant points of a chapter in a short amount of time.
• The solutions also prepare you for managing your stress and time, giving you a better chance of finishing your examinations on time.
• NCERT solutions class 12 mathematics chapter 8 will assist you in grasping the fundamental concepts quickly and easily.

NCERT Exemplar Class 12 Mathematics

Here are some solutions and problems to help students prepare for their final exams. These example questions are a little more complex, and they cover a variety of concepts in each chapter of Class 12 Mathematics subject. Students will fully understand all the concepts covered in each chapter by practising these NCERT Exemplars for mathematics Class 12. Each question in these materials is connected to concepts covered in the CBSE Class 12 syllabus (2022-2023). Our experts provide the best solutions to challenges that students confront. To match the ideas taught in each class and provide the greatest practising materials or worksheets for students, all these questions reflect the question pattern found in NCERT books.

Q.1 Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.

Ans.

The area of the region bounded by the curve, y2 = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

$\begin{array}{l}Area\text{of ABCD}={\int }_{1}^{4}ydx\\ \text{}\text{}\text{}={\int }_{1}^{4}\sqrt{x}\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{}={\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{1}^{4}=\frac{2}{3}\left[{4}^{\frac{3}{2}}\text{\hspace{0.17em}}-{1}^{\frac{3}{2}}\right]\\ \text{}\text{}\text{}=\frac{2}{3}\left(8-1\right)\\ \text{}\text{}\text{}=\frac{14}{3}\text{\hspace{0.17em}}units\end{array}$

Q.2 Find the area of the region bounded by y2= 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Ans.

The area of the region bounded by the curve, y2 = 9x, the lines, x = 2 and x = 4, and the x-axis is the area ABCD.

$\begin{array}{l}Area\text{of ABCD}={\int }_{2}^{4}ydx\\ \text{}\text{}\text{}={\int }_{2}^{4}3\sqrt{x}\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{}=3{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{2}^{4}=3×\frac{2}{3}\left[{4}^{\frac{3}{2}}\text{\hspace{0.17em}}-{2}^{\frac{3}{2}}\right]\\ \text{}\text{}\text{}=2\left(8-2\sqrt{2}\right)\\ \text{}\text{}\text{}=\left(16-4\sqrt{2}\right)\text{\hspace{0.17em}}units\end{array}$

Q.3 Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Ans.

The area of the region bounded by the curve, x2 = 4y, the lines, y = 2 and y = 4, and the y-axis is the area ABCD.

$\begin{array}{l}Area\text{of ABCD}={\int }_{2}^{4}xdy\\ \text{}\text{}\text{}={\int }_{2}^{4}2\sqrt{y}\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{}=2{\left[\frac{{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{2}^{4}=\frac{4}{3}\left[{4}^{\frac{3}{2}}\text{\hspace{0.17em}}-{2}^{\frac{3}{2}}\right]\\ \text{}\text{}\text{}=\frac{4}{3}\left(8-2\sqrt{2}\right)\\ \text{}\text{}\text{}=\left(\frac{32-8\sqrt{2}}{3}\right)\text{\hspace{0.17em}}units\end{array}$

Q.4 Find the area of the region bounded by the ellipse (x2/16) + (y2/9) = 1.

Ans.

The given ellipse (x2/16) + (y2/9)=1 can be represented as given below:

Ellipse is symmetrical about x-axis and y-axis.
So, area bounded by ellipse = 4 x Area of OAB

$\begin{array}{l}Area\text{of OAB}={\int }_{0}^{4}ydx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{4}3\sqrt{1-\frac{{x}^{2}}{16}}dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}{\int }_{0}^{4}\sqrt{16-{x}^{2}}dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}{\left[\frac{x}{2}\sqrt{16-{x}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\frac{x}{4}\right]}_{0}^{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}\left[\frac{4}{2}\sqrt{16-{4}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\frac{4}{4}\right]-\frac{3}{4}\left[\frac{0}{2}\sqrt{16-{0}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\frac{0}{4}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}×8\frac{\pi }{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3\pi \text{\hspace{0.17em}}square\text{\hspace{0.17em}}units\\ Therefore,\text{area bounded by ellipse}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4×Area\text{of OAB}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{4}×3\pi \text{\hspace{0.17em}}square\text{\hspace{0.17em}}units\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}square\text{\hspace{0.17em}}units\end{array}$

Q.5 Find the area of the region bounded by the ellipse (x2/4) + (y2/9) = 1.

Ans.

The given ellipse (x2/4) + (y2/9) =1 can be represented as given below:

$\begin{array}{l}The\text{given equation of ellipse is}\\ \frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1\\ \text{}y=3\sqrt{1-\frac{{x}^{2}}{4}}\\ It\text{can be observed that ellipse is symmetrical about x-axis and}\\ \text{y-axis}\text{.}\\ \therefore \text{Area bounded by ellipse}=4×Area\text{OAB}\\ \therefore \text{Area of OAB}={\int }_{0}^{2}ydx\\ \text{}\text{}\text{}={\int }_{0}^{2}3\sqrt{1-\frac{{x}^{2}}{4}}\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{}=\frac{3}{2}{\int }_{0}^{2}\sqrt{\left(4-{x}^{2}\right)}\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{}=\frac{3}{2}{\left[\frac{x}{2}\sqrt{\left(4-{x}^{2}\right)}+\frac{4}{2}{\mathrm{sin}}^{-1}\frac{x}{2}\right]}_{0}^{2}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}=\frac{3}{2}\left[\frac{2}{2}\sqrt{\left(4-{2}^{2}\right)}+\frac{4}{2}{\mathrm{sin}}^{-1}\frac{2}{2}\right]-\frac{3}{2}\left[\frac{0}{2}\sqrt{\left(4-{0}^{2}\right)}+\frac{4}{2}{\mathrm{sin}}^{-1}\frac{0}{2}\right]\\ \text{}\text{}\text{}=\frac{3}{2}×2\frac{\pi }{2}\\ \text{}\text{}\text{}=\frac{3}{2}\text{​}\pi \text{\hspace{0.17em}}square\text{units}\\ \text{Therefore, area bounded by the ellipse}\\ \text{}\text{}\text{}=4×\frac{3}{2}\text{​}\pi \text{\hspace{0.17em}}square\text{units}\\ \text{}\text{}\text{}=\text{\hspace{0.17em}}\text{6}\text{\hspace{0.17em}}square\text{unit}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Findtheareaoftheregioninthefirst} \mathrm{quadrantenclosed}\\ \mathrm{byx}–\mathrm{axis},\mathrm{linex}=\sqrt{3}{\mathrm{yandthecirclex}}^{2}+{\mathrm{y}}^{2}=4.\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{circle},{\mathrm{x}}^{2}+{\mathrm{y}}^{\mathrm{2}}={\mathrm{a}}^{\mathrm{2}},\mathrm{ }\mathrm{cut}\mathrm{ }\mathrm{off}\mathrm{ }\mathrm{by}\\ \mathrm{the}\mathrm{line},\mathrm{x}=\frac{\mathrm{a}}{\sqrt{2}},\mathrm{is}\mathrm{the}\mathrm{area}\mathrm{ABCDA}\mathrm{.}\end{array}$

$\begin{array}{l}It\text{is observed that the area ABCD is symmetrical about x-axis}\text{.}\\ \therefore \text{area ABCD}=2×Area\text{ABC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{of}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ABC}={\int }_{\frac{a}{\sqrt{2}}}^{a}ydx\\ \text{}\text{}\text{}={\int }_{\frac{a}{\sqrt{2}}}^{a}\sqrt{{a}^{2}-{x}^{2}}dx\\ \text{}\text{}\text{}={\left[\frac{x}{2}\sqrt{{a}^{2}-{x}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\frac{x}{a}\right]}_{\frac{a}{\sqrt{2}}}^{a}\\ \text{}=\left[\frac{a}{2}\sqrt{{a}^{2}-{a}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\frac{a}{a}\right]-\left[\frac{\left(\frac{a}{\sqrt{2}}\right)}{2}\sqrt{{a}^{2}-{\left(\frac{a}{\sqrt{2}}\right)}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\frac{\left(\frac{a}{\sqrt{2}}\right)}{a}\right]\\ \text{}\text{}\text{}=\frac{{a}^{2}}{2}×\frac{\pi }{2}-\frac{a}{2\sqrt{2}}×\frac{a}{\sqrt{2}}-\frac{{a}^{2}}{2}\left(\frac{\pi }{4}\right)\\ \text{}\text{}\text{}=\frac{{a}^{2}}{2}×\frac{\pi }{2}-\frac{{a}^{2}}{4}-\frac{{a}^{2}}{2}\left(\frac{\pi }{4}\right)\\ \text{}\text{}\text{}=\frac{{a}^{2}}{4}\left(\pi -1-\frac{\pi }{2}\right)\\ \text{}\text{}\text{}=\frac{{a}^{2}}{4}\left(\frac{\pi }{2}-1\right)\\ \therefore Area\text{ABCD}=2\left\{\frac{{a}^{2}}{4}\left(\frac{\pi }{2}-1\right)\right\}\\ \text{}\text{}\text{}=\frac{{a}^{2}}{2}\left(\frac{\pi }{2}-1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}square\text{\hspace{0.17em}}units\\ Therefore,{\text{the area of the smaller part of the circle, x}}^{\text{2}}{\text{+y}}^{\text{2}}={\text{a}}^{\text{2}},\text{}\\ \text{cut off by the line, x=}\frac{a}{\sqrt{2}}\text{is}\frac{{a}^{2}}{2}\left(\frac{\pi }{2}-1\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}square\text{\hspace{0.17em}}units.\end{array}$

Q.7 The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.

Ans.

The line x = a divides the area bounded by the parabola and line x= 4 into two equal parts.

$\begin{array}{l}Area\text{\hspace{0.17em}}OED={\int }_{0}^{a}ydx\\ \text{}\text{}={\int }_{0}^{a}\sqrt{x}dx\\ \text{}\text{}={\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{a}\\ \text{}\text{}=\frac{2}{3}{a}^{\frac{3}{2}}\\ Area\text{​}\text{of CDEF}\\ \text{}\text{}={\int }_{a}^{4}ydx\\ \text{}\text{}={\int }_{a}^{4}\sqrt{x}dx\\ \text{}\text{}={\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{a}^{4}\\ \text{}\text{}=\frac{2}{3}\left({4}^{\frac{3}{2}}-{a}^{\frac{3}{2}}\right)\\ \text{}\text{}=\frac{2}{3}\left(8-{a}^{\frac{3}{2}}\right)\\ According\text{to equation}\left(i\right),\text{we have}\\ \text{}\frac{2}{3}{a}^{\frac{3}{2}}=\frac{2}{3}\left(8-{a}^{\frac{3}{2}}\right)\\ ⇒\text{}{a}^{\frac{3}{2}}=\left(8-{a}^{\frac{3}{2}}\right)\\ ⇒2{a}^{\frac{3}{2}}=8\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}^{\frac{3}{2}}=4\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a={4}^{\frac{2}{3}}\\ Therefore,\text{the value of a is}{4}^{\frac{2}{3}}.\end{array}$

Q.8 Find the area of the region bounded by the parabola y = x2 and y = |x|.

Ans.

The area bounded by the parabola y = x2 and
y = |x| is shown as below:

Since the required area is symmetrical about y-axis.

So, area OACO = area OBDO

The points of intersection of parabola and line are A(1,1) and B(–1, 1).

$\begin{array}{l}Area\text{of OACO}=Area\text{}\Delta \text{AMO}-area\text{\hspace{0.17em}}OMACO\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}×OM×AM-{\int }_{0}^{1}ydx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}×1×1-{\int }_{0}^{1}{x}^{2}dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}-{\left[\frac{{x}^{3}}{3}\right]}_{0}^{1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}-\left[\frac{{1}^{3}}{3}-\frac{{0}^{3}}{3}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}-\frac{1}{3}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{6}\\ \therefore \text{Required area}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(Area\text{of OACO}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2×\frac{1}{6}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3}square\text{units}\end{array}$

Q.9 Find the area of the region bounded by the curve x2 = 4y and the line x = 4y – 2.

Ans.

The area bounded by the parabola x2 = 4y and x = 4y – 2 is shown in figure.
The intersection points of line and parabola are A(–1, 1/4) and B(2,1).

$\begin{array}{l}Here\text{\hspace{0.17em}}\text{\hspace{0.17em}}AL\perp X-axis\text{and BM}\perp X-axis.\text{\hspace{0.17em}}Then\\ Area\text{\hspace{0.17em}}OBAO=Area\text{\hspace{0.17em}}OBCO+Area\text{\hspace{0.17em}}OACO\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(Area\text{\hspace{0.17em}}OMBCO-Area\text{\hspace{0.17em}}OMBO\right)+\left(Area\text{\hspace{0.17em}}OLAC-Area\text{\hspace{0.17em}}OLAO\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({\int }_{0}^{2}\frac{x+2}{4}dx-{\int }_{0}^{2}\frac{{x}^{2}}{4}dx\right)+\left({\int }_{-1}^{0}\frac{x+2}{4}dx-{\int }_{-1}^{0}\frac{{x}^{2}}{4}dx\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}{\left[\frac{{x}^{2}}{2}+2x\right]}_{0}^{2}-\frac{1}{4}{\left[\frac{{x}^{3}}{3}\right]}_{0}^{2}+\frac{1}{4}{\left[\frac{{x}^{2}}{2}+2x\right]}_{-1}^{0}-\frac{1}{4}{\left[\frac{{x}^{3}}{3}\right]}_{-1}^{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}\left[\frac{{2}^{2}}{2}+2\left(2\right)\right]-\frac{1}{4}\left[\frac{{2}^{3}}{3}\right]+0-\frac{1}{4}\left[\frac{{\left(-1\right)}^{2}}{2}+2\left(-1\right)\right]-0+\frac{1}{4}\left[\frac{{\left(-1\right)}^{3}}{3}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}×6-\frac{1}{4}×\frac{8}{3}-\frac{1}{4}×\frac{-3}{2}+\frac{1}{4}×\frac{-1}{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}-\frac{2}{3}+\frac{3}{8}-\frac{1}{12}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{36-16+9-2}{24}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{27}{24}\text{\hspace{0.17em}}=\frac{9}{8}\text{\hspace{0.17em}}sq\text{\hspace{0.17em}}units\\ Thus,\text{the area of OBAO is}\frac{9}{8}\text{\hspace{0.17em}}sq.\text{\hspace{0.17em}}\text{\hspace{0.17em}}units.\end{array}$

Q.10 Find the area of the region bounded by the curve y2 = 4x and the line x = 3.

Ans.

The region bounded by the parabola and line is given below:

The area COAC is symmetrical about x-axis. So,
Area OABO = Area OCBO
Then, Area OCAO = 2(area OABO)

$\begin{array}{l}Area\text{\hspace{0.17em}}OACO=2{\int }_{0}^{3}ydx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2{\int }_{0}^{3}2\sqrt{x}dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{3}×4\left[{3}^{\frac{3}{2}}-0\right]\text{\hspace{0.17em}}=8\sqrt{3}\text{\hspace{0.17em}}\\ Therefore,\text{the required area is 8}\sqrt{3}\text{\hspace{0.17em}}square\text{\hspace{0.17em}}units.\end{array}$

Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is

$\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(b\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(c\right)\text{\hspace{0.17em}}\frac{\pi }{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(d\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\pi }{4}$

Ans.

The area bounded by the circle and the lines x = 0 and x = 2 lies in first quadrant.

$\begin{array}{l}\therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}OABO={\int }_{0}^{2}ydx\\ \text{}\text{}\text{}={\int }_{0}^{2}\sqrt{4-{x}^{2}}dx\\ \text{}\text{}\text{}={\left[\frac{x}{2}\sqrt{4-{x}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{2}\right)\right]}_{0}^{2}\\ \text{}\text{}\text{}=\left[\frac{2}{2}\sqrt{4-{2}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{2}{2}\right)\right]-\left[\frac{0}{2}\sqrt{4-{0}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{0}{2}\right)\right]\\ \text{}\text{}\text{}=\left[0+2×\frac{\pi }{2}\right]-\left[0+0\right]=\pi \\ Thus,\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}the\text{correct option is A}\text{.}\end{array}$

Area of the region bounded by the curve y2 = 4x, and y-axis and the line y = 3 is

(A) 2 (B) 9/4 (C) 9/3 (D) 9/2

Ans.

The area bounded by the parabola and the lines y = 0 and y = 3 lies in first quadrant.

$\begin{array}{l}\therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}OAB={\int }_{0}^{3}xdy\\ \text{}\text{}\text{}={\int }_{0}^{2}\frac{{y}^{2}}{4}dy\\ \text{}\text{}\text{}=\frac{1}{4}{\left[\frac{{y}^{3}}{3}\right]}_{0}^{3}\\ \text{}\text{}\text{}=\frac{1}{12}\left[{3}^{3}-{0}^{3}\right]\\ \text{}\text{}\text{}=\frac{27}{12}=\frac{9}{4}\\ Thus,\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}the\text{correct option is B}\text{.}\end{array}$

Q.13 Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.

Ans.

The required area is represented by shaded area OBCDO.

$\begin{array}{l}On{\text{solving the given equations 4x}}^{\text{2}}{\text{+ 4y}}^{\text{2}}={\text{9 and parabola x}}^{2}=\text{4y,}\\ we\text{get the coordinates of point of intersection as B}\left(\sqrt{2},\text{\hspace{0.17em}}\frac{1}{2}\right)\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\\ C\left(-\sqrt{2},\text{\hspace{0.17em}}\frac{1}{2}\right).This\text{area is symmetrical about y-axis}\text{.}\\ \therefore \text{​}\text{\hspace{0.17em}}AreaOBCD=2\left(area\text{\hspace{0.17em}}OBCO\right)\\ Since,\text{\hspace{0.17em}}BM\perp X-axis.\\ So,\text{\hspace{0.17em}}the\text{coordinates of M are}\left(\sqrt{2},0\right).\\ Therefore,\text{\hspace{0.17em}}area\text{OBCO}=Area\text{\hspace{0.17em}}OMBCO-Area\text{\hspace{0.17em}}OMBO\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\sqrt{2}}\sqrt{\frac{9-4{x}^{2}}{4}}dx-{\int }_{0}^{\sqrt{2}}\frac{{x}^{2}}{4}dx\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}{\int }_{0}^{\sqrt{2}}\sqrt{9-4{x}^{2}}dx-{\int }_{0}^{\sqrt{2}}\frac{{x}^{2}}{4}dx\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}{\int }_{0}^{2\sqrt{2}}\sqrt{9-{t}^{2}}\frac{dt}{2}-\frac{1}{4}{\int }_{0}^{\sqrt{2}}{x}^{2}dx\text{}\left[\begin{array}{l}Let\text{t}=2x⇒\frac{dt}{dx}=2\\ and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t=2×\sqrt{2},\left(\text{\hspace{0.17em}}When\text{\hspace{0.17em}}x=\sqrt{2}\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\sqrt{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t=2×0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{\hspace{0.17em}}When\text{\hspace{0.17em}}x=0\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0\end{array}\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}{\left[\frac{t}{2}\sqrt{9-{t}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{t}{3}\right]}_{0}^{2\sqrt{2}}-\frac{1}{4}{\left[\frac{{x}^{3}}{3}\right]}_{0}^{\sqrt{2}}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}\left[\frac{2\sqrt{2}}{2}\sqrt{9-{\left(2\sqrt{2}\right)}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right]-\\ \text{}\text{}\text{}\text{}\frac{1}{4}\left[\left(0\right)\sqrt{9-4{\left(0\right)}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{2\left(0\right)}{3}\right]-\frac{1}{4}\left[\frac{{\left(\sqrt{2}\right)}^{3}}{3}\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{4}\left(\sqrt{2}+\frac{9}{2}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right)-\frac{\sqrt{2}}{6}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{2}}{4}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}-\frac{\sqrt{2}}{6}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{2}}{12}+\frac{9}{8}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right)\\ Therefore,\text{area of shaded region is 2}×\frac{1}{2}\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\mathrm{sin}}^{-1}\frac{2\sqrt{2}}{3}\right)\text{\hspace{0.17em}}square\text{​}\text{units}\text{.}\end{array}$

Q.14 Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1.

Ans.

The required area bounded by (x – 1)2 + y2 = 1 and x2 + y2 = 1 is represented by shaded area ACBOA.

$\begin{array}{l}On\text{solving the given equations}{\left(\text{x}-\text{1}\right)}^{\text{2}}{\text{+ y}}^{\text{2}}={\text{1 and x}}^{\text{2}}{\text{+y}}^{\text{2}}=1,\\ we\text{get the coordinates of point of intersection as A}\left(\frac{1}{2},\text{\hspace{0.17em}}\frac{\sqrt{3}}{2}\right)\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\\ B\left(\frac{1}{2},-\text{\hspace{0.17em}}\frac{\sqrt{3}}{2}\right).This\text{area is symmetrical about x-axis}\text{.}\\ \therefore \text{​}\text{\hspace{0.17em}}AreaOBCAO=2\left(area\text{\hspace{0.17em}}OCAO\right)\\ Join\text{​}\text{AB such that}\text{\hspace{0.17em}}AM\perp OC.\\ So,\text{\hspace{0.17em}}the\text{coordinates of M are}\left(\frac{1}{2},0\right).\\ Therefore,\text{\hspace{0.17em}}area\text{OCAO}=Area\text{\hspace{0.17em}}OMAO+Area\text{\hspace{0.17em}}MCAM\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\frac{1}{2}}\sqrt{1-{\left(x-1\right)}^{2}}dx+{\int }_{\frac{1}{2}}^{1}\sqrt{1-{x}^{2}}dx\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[\frac{\left(x-1\right)}{2}\sqrt{1-{\left(x-1\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(x-1\right)\right]}_{0}^{\frac{1}{2}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}+{\left[\frac{x}{2}\sqrt{1-{x}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}x\right]}_{\frac{1}{2}}^{1}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\frac{\left(\frac{1}{2}-1\right)}{2}\sqrt{1-{\left(\frac{1}{2}-1\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{1}{2}-1\right)\right]\\ \text{}-\left[\frac{\left(0-1\right)}{2}\sqrt{1-{\left(0-1\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(0-1\right)\right]+\left[\frac{1}{2}\sqrt{1-{1}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}1\right]\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}-\left[\frac{\frac{1}{2}}{2}\sqrt{1-{\left(\frac{1}{2}\right)}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[-\frac{\sqrt{3}}{8}+\frac{1}{2}\left(-\frac{\pi }{6}\right)-\frac{1}{2}\left(-\frac{\pi }{2}\right)\right]+\frac{1}{2}\left(\frac{\pi }{2}\right)-\frac{\sqrt{3}}{8}-\frac{1}{2}\left(\frac{\pi }{6}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{\sqrt{3}}{8}-\frac{\pi }{12}+\frac{\pi }{4}+\frac{\pi }{4}-\frac{\sqrt{3}}{8}-\frac{\pi }{12}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\\ Therefore,\text{required area of OBCAO}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\frac{2\pi }{6}-\frac{\sqrt{3}}{4}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)\text{\hspace{0.17em}}\text{square units}\end{array}$

Q.15 Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3.

Ans.

The area bounded by y = x2 +2, y = x, x = 0 and x = 3 is represented by the shaded area

$\begin{array}{l}Then,\text{\hspace{0.17em}}area\text{\hspace{0.17em}}OCBAO=area\text{\hspace{0.17em}}ODBAO-area\text{\hspace{0.17em}}ODCO\\ \text{}\text{}\text{}\text{}={\int }_{0}^{3}\left({x}^{2}+2\right)dx-{\int }_{0}^{3}x\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{}\text{}={\left[\frac{{x}^{3}}{3}+2x\right]}_{0}^{3}-{\left[\frac{{x}^{2}}{2}\right]}_{0}^{3}\\ \text{}\text{}\text{}\text{}=\left[\frac{{3}^{3}}{3}+2\left(3\right)-0-0\right]-\left[\frac{{3}^{2}}{2}-0\right]\\ \text{}\text{}\text{}\text{}=15-\frac{9}{2}=\frac{21}{2}\text{\hspace{0.17em}}units\\ Thus,\text{the required area of shaded region is}\frac{21}{2}\text{\hspace{0.17em}}sq\text{\hspace{0.17em}}units.\end{array}$

Q.16 Using integration find the area of region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).

Ans.

Here, BL and CM are perpendicular to x-axis.

$\begin{array}{l}Equation\text{of line AB is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}-0=\frac{3-0}{1+1}\left(x+1\right)\\ \text{}y=\frac{3}{2}\left(x+1\right)\\ Equation\text{of line BC is}\\ \text{y}-3=\frac{2-3}{3-1}\left(x-1\right)\\ \text{}y=\frac{-1}{2}\left(x-1\right)+3\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(-x+7\right)\\ Equation\text{of line CA is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}-2=\frac{2-0}{3+1}\left(x-3\right)\\ \text{}y=\frac{1}{2}\left(x-3\right)+2\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(x+1\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}\Delta ACB=Area\left(ALBA\right)+Area\left(BLMCB\right)-Area\left(\Delta AMCA\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{-1}^{1}\frac{3}{2}\left(x+1\right)dx+{\int }_{1}^{3}\frac{1}{2}\left(-x+7\right)dx-{\int }_{-1}^{3}\frac{1}{2}\left(x+1\right)dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}{\left[\frac{{x}^{2}}{2}+x\right]}_{-1}^{1}+\frac{1}{2}{\left[\frac{-{x}^{2}}{2}+7x\right]}_{1}^{3}-\frac{1}{2}{\left[\frac{{x}^{2}}{2}+x\right]}_{-1}^{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}\left[\frac{{1}^{2}}{2}+1-\frac{{\left(-1\right)}^{2}}{2}-\left(-1\right)\right]+\frac{1}{2}\left[\frac{-{3}^{2}}{2}+7\left(3\right)+\frac{{1}^{2}}{2}-7\left(1\right)\right]\\ \text{}\text{}\text{}\text{}\text{}-\frac{1}{2}\left[\frac{{3}^{2}}{2}+3-\frac{{\left(-1\right)}^{2}}{2}-\left(-1\right)\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{2}\left(2\right)+\frac{1}{2}\left(-\frac{9}{2}+21+\frac{1}{2}-7\right)-\frac{1}{2}\left(\frac{9}{2}+3-\frac{1}{2}+1\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3+5-4\\ \end{array}$

=4square units

Q.17 Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.

Ans.

The coordinates of intersection points of the given lines are A(0, 1), B(4, 13) and C(4, 9).

$\begin{array}{l}Area\text{\hspace{0.17em}}\Delta ABC=Area\text{\hspace{0.17em}}OLBAO-Area\text{\hspace{0.17em}}\text{\hspace{0.17em}}OLCAO\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{4}\left(3x+1\right)dx-{\int }_{0}^{4}\left(2x+1\right)dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[3\frac{{x}^{2}}{2}+x\right]}_{0}^{4}-{\left[2.\frac{{x}^{2}}{2}+x\right]}_{0}^{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[3\frac{{\left(4\right)}^{2}}{2}+4-3\frac{{\left(0\right)}^{2}}{2}-0\right]-\left[2.\frac{{4}^{2}}{2}+4-2.\frac{{0}^{2}}{2}+0\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(24+4\right)-\left(16+4\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=28-20\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8\text{square units}\end{array}$

Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y =2 is

$\begin{array}{l}\left(a\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(\pi -2\right)\\ \left(b\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\pi -2\\ \left(c\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\pi -1\\ \left(d\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(\pi +2\right)\end{array}$

Ans.

The smaller area enclosed by the circle, x2 + y2 = 4 and line x + y = 2 is given by shaded region ABCA.

$\begin{array}{l}Area\text{\hspace{0.17em}}ABCA=Area\text{\hspace{0.17em}}OACBO-Area\Delta AOB\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{2}\sqrt{\left(4-{x}^{2}\right)}dx-{\int }_{0}^{2}\left(2-x\right)dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[\frac{x}{2}\sqrt{4-{x}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{2}\right)\right]}_{0}^{2}-{\left[2x-\frac{{x}^{2}}{2}\right]}_{0}^{2}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\frac{2}{2}\sqrt{4-{2}^{2}}+2{\mathrm{sin}}^{-1}\left(\frac{2}{2}\right)\right]-\left[2\left(2\right)-\frac{{2}^{2}}{2}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(2×\frac{\pi }{2}\right)-\left(4-2\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(\pi -2\right)\text{square units}\\ \text{Therefore, correct option is B}\text{.}\end{array}$

Q.19 Area lying between the curves y2 = 4x and y = 2x is
(A) 2/3 (B) 1/3 (C) 1/4 (D) 3/4

Ans.

The area lying between y2 = 4x and y = 2x is shown by shaded area given graph.

$\begin{array}{l}Area\text{\hspace{0.17em}}OBAO=area\text{\hspace{0.17em}}OCABO-area\text{\hspace{0.17em}}OCAO\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{1}2\sqrt{x}dx-{\int }_{0}^{1}2xdx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{1}-2{\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}\left[1-0\right]-\left[{1}^{2}-0\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}-1\text{\hspace{0.17em}}=\frac{1}{3}\text{\hspace{0.17em}}square\text{units}\\ \end{array}$

Therefore, correct option is B.

Q.20 Find the area under the given curves and given lines:
(i) y = x2, x = 1, x = 2 and x- axis

(ii) y = x4, x = 1, x = 5 and x- axis

Ans.

(i) The required area ADCBA is represented by shaded region in graph.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}ABCDA={\int }_{1}^{2}{x}^{2}dx\\ \text{}\text{}\text{}={\left[\frac{{x}^{3}}{3}\right]}_{1}^{2}\\ \text{}\text{}\text{}=\frac{1}{3}\left[{2}^{3}-{1}^{3}\right]\\ \text{}\text{}\text{}=\frac{7}{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}s\text{quare units}\end{array}$

(ii) The required area ADCBA is represented by shaded region in graph.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\text{\hspace{0.17em}}ABCDA={\int }_{1}^{5}{x}^{4}dx\\ \text{}\text{}\text{}={\left[\frac{{x}^{5}}{5}\right]}_{1}^{5}\\ \text{}\text{}\text{}=\frac{1}{5}\left[{5}^{5}-{1}^{5}\right]\\ \text{}\text{}\text{}=\frac{3124}{5}\\ \text{}\text{}\text{}=624.8\text{\hspace{0.17em}}\text{\hspace{0.17em}}s\text{quare units}\end{array}$

Q.21 Find the area between the curves y = x and y = x2.

Ans.

The point of intersection of curve y = x2 and line y = x is (1, 1). Here AC is drawn as perpendicular to x-axis.

$\begin{array}{l}\therefore Area\text{\hspace{0.17em}}OBAO=Area\left(\Delta OCA\right)-Area\left(OCABO\right)\\ \text{}\text{}\text{}={\int }_{0}^{1}xdx-{\int }_{0}^{1}{x}^{2}dx\\ \text{}\text{}\text{}={\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}-{\left[\frac{{x}^{3}}{3}\right]}_{0}^{1}\\ \text{}\text{}\text{}=\left[\frac{{1}^{2}}{2}-0\right]-\left[\frac{{1}^{3}}{3}-0\right]\\ \text{}\text{}\text{}=\frac{1}{2}-\frac{1}{3}\\ \text{}\text{}\text{}=\frac{1}{6}\text{square units}\end{array}$

Q.22 Find the area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4.

Ans.

The area of the region lying in the first quadrant and bounded by y = 4x2, x = 0, y = 1 and y = 4 is shown by shaded region ABCDA.

$\begin{array}{l}\therefore Area\text{\hspace{0.17em}}ABCD={\int }_{1}^{4}\frac{\sqrt{y}}{2}dy\text{}\left[\because y=4{x}^{2}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}{\int }_{1}^{4}{y}^{\frac{1}{2}}dy\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}{\left[\frac{{y}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{1}^{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3}\left[{4}^{\frac{3}{2}}-{1}^{\frac{3}{2}}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3}\left(8-1\right)\text{\hspace{0.17em}}=\frac{7}{3}\text{square units}\end{array}$

Q.23

$Sketch\text{}the\text{}graph\text{}of\text{}y=|x+3|\text{}and\text{}evaluate\text{\hspace{0.17em}}{\int }_{-6}^{0}|x+3|\text{​}dx.$

Ans.

The graph of y = |x + 3| is as shown below:

$\begin{array}{l}\because \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=|x+3|=\left\{\begin{array}{l}-\left(x+3\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+3<0⇒x<-3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x+3\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+3\ge 0⇒x\ge -3\end{array}\\ \therefore {\int }_{-6}^{0}|\left(x+3\right)|dx={\int }_{-6}^{-3}-\left(x+3\right)dx+{\int }_{-3}^{0}\left(x+3\right)dx\\ \text{}\text{}\text{}=-{\left[\frac{{x}^{2}}{2}+3x\right]}_{-6}^{-3}+{\left[\frac{{x}^{2}}{2}+3x\right]}_{-3}^{0}\\ \text{}\text{}\text{}=-\left[\frac{{\left(-3\right)}^{2}}{2}+3\left(-3\right)-\frac{{\left(-6\right)}^{2}}{2}-3\left(-6\right)\right]+\left[0-\frac{{\left(-3\right)}^{2}}{2}-3\left(-3\right)\right]\\ \text{}\text{}\text{}=-\left(\frac{9}{2}-9-18+18\right)+\left(-\frac{9}{2}+9\right)\\ \text{}\text{}\text{}=\frac{9}{2}+\frac{9}{2}=9\text{square units}\end{array}$

Q.24

$\begin{array}{l}\mathrm{Findtheareaboundedbythecurvey}=\mathrm{sinxbetweenx}=0\\ \mathrm{andx}=2\mathrm{\pi }.\end{array}$

Ans.

The graph of y = sin x is shown below:

$\begin{array}{l}\mathrm{Required}\mathrm{area}=\mathrm{Area}\mathrm{OABO}+\mathrm{Area}\mathrm{BCDB}\\ ={\int }_{0}^{\mathrm{\pi }}\mathrm{sinxdx}+|{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\mathrm{sinxdx}|\\ ={\left[-\mathrm{cosx}\right]}_{0}^{\mathrm{\pi }}+|{\left[-\mathrm{cosx}\right]}_{\mathrm{\pi }}^{2\mathrm{\pi }}|\\ =-\left[\mathrm{cos\pi }+\mathrm{cos}0\right]+|\mathrm{cos}2\mathrm{\pi }-\mathrm{cos\pi }|\\ =-\left[-1-1\right]+\left(1+1\right)\\ \mathrm{ }=4\mathrm{ }\mathrm{square}\mathrm{units}\end{array}$

Q.25 Find the area enclosed between the parabola y2 = 4ax and the line y = mx.

Ans.

Solving y2 = 4ax and y = mx, we get x = (4a/m2)
and y = (4a/m).

$\begin{array}{l}The\text{points of intersection of both the curves are}\left(0,0\right)\text{and}\\ \left(\frac{4a}{{m}^{2}},\text{\hspace{0.17em}}\frac{4a}{m}\right).\\ \therefore Area\text{\hspace{0.17em}}OBAO=Area\text{\hspace{0.17em}}OCABO-Area\text{\hspace{0.17em}}\left(\Delta OCA\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\frac{4a}{{m}^{2}}}2\sqrt{ax}dx-{\int }_{0}^{\frac{4a}{{m}^{2}}}mx\text{\hspace{0.17em}}dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\sqrt{a}{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{\frac{4a}{{m}^{2}}}-m{\left[\frac{{x}^{2}}{2}\right]}_{0}^{\frac{4a}{{m}^{2}}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}\sqrt{a}\left\{{\left(\frac{4a}{{m}^{2}}\right)}^{\frac{3}{2}}-0\right\}-\frac{m}{2}\left\{{\left(\frac{4a}{{m}^{2}}\right)}^{2}-0\right\}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}\sqrt{a}\left(\frac{8a\sqrt{a}}{{m}^{3}}\right)-\frac{m}{2}\left(\frac{16{a}^{2}}{{m}^{4}}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}×\frac{8{a}^{2}}{{m}^{3}}-\frac{8{a}^{2}}{{m}^{3}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{8{a}^{2}}{3{m}^{3}}\end{array}$

Q.26 Find the area enclosed between the parabola 4y = 3x2 and the line 2y = 3x + 12.

Ans.

The area enclosed between parabola 4y = 3x2 and line 2y = 3x + 12 is shown by shaded part AOBA in figure and point of intersection of the given curves are A(–2, 3) and B(4, 12). Two perpendiculars AC and BD are drawn on x-axis.

$\begin{array}{l}\therefore Area\text{\hspace{0.17em}}\text{AOBA}=Area\text{\hspace{0.17em}}CDBA-\left(Area\text{\hspace{0.17em}}ACOA+Area\text{\hspace{0.17em}}ODBO\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{-2}^{4}\frac{3x+12}{2}dx-{\int }_{-2}^{4}\frac{3{x}^{2}}{4}dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}{\left[3\frac{{x}^{2}}{2}+12x\right]}_{-2}^{4}-\frac{3}{4}{\left[\frac{{x}^{3}}{3}\right]}_{-2}^{4}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left\{\frac{3}{2}×{4}^{2}+12×4-\frac{3}{2}×{\left(-2\right)}^{2}-12×\left(-2\right)\right\}\\ \text{}\text{}\text{}\text{}\text{}\text{}-\frac{1}{4}\left\{{4}^{3}-{\left(-2\right)}^{3}\right\}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(24+48-6+24\right)-\frac{1}{4}\left(64+8\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}×90-18\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=45-18\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=27\text{\hspace{0.17em}}square\text{\hspace{0.17em}}\text{\hspace{0.17em}}units\end{array}$

Q.27

$\begin{array}{l}Find\text{}the\text{}area\text{}of\text{}the\text{}smaller\text{}region\text{}bounded\text{}by\text{}the\text{}ellipse\\ \frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1\text{​}andtheline\frac{x}{3}+\frac{y}{2}=1.\end{array}$

Ans.

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

$\begin{array}{l}\therefore Area\left(ABCA\right)=Area\left(OBCAO\right)-Area\left(OBAO\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{3}2\sqrt{1-\frac{{x}^{2}}{9}}dx-{\int }_{0}^{3}2\left(1-\frac{x}{3}\right)dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{3}{\int }_{0}^{3}\sqrt{9-{x}^{2}}dx-\frac{2}{3}{\int }_{0}^{3}\left(3-x\right)dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{3}{\left[\frac{x}{2}\sqrt{9-{x}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{3}\right)\right]}_{0}^{3}-\frac{2}{3}{\left[3x-\frac{{x}^{2}}{2}\right]}_{0}^{3}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{3}\left\{\frac{3}{2}\sqrt{9-{3}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{3}{3}\right)-\frac{0}{2}\sqrt{9-{0}^{2}}-\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{0}{3}\right)\right\}\\ \text{}\text{}\text{}\text{}-\frac{2}{3}\left\{3\left(3\right)-\frac{{3}^{2}}{2}-3\left(0\right)+\frac{{0}^{2}}{2}\right\}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{3}\left\{\frac{9}{2}×\frac{\pi }{2}\right\}-\frac{2}{3}\left\{9-\frac{9}{2}\right\}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3\pi }{2}-3=\frac{3}{2}\left(\pi -2\right)\text{\hspace{0.17em}}square\text{units}\end{array}$

Q.28

$\begin{array}{l}Find\text{}the\text{}area\text{}of\text{}the\text{}smaller\text{}region\text{}bounded\text{}by\text{}the\text{}ellipse\\ \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\text{​}andtheline\frac{x}{a}+\frac{y}{b}=1.\end{array}$

Ans.

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

$\begin{array}{l}\therefore Area\left(ABCA\right)=Area\left(OBCAO\right)-Area\left(OBAO\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{a}b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}dx-{\int }_{0}^{a}b\left(1-\frac{x}{a}\right)dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{b}{a}{\int }_{0}^{a}\sqrt{{a}^{2}-{x}^{2}}dx-\frac{b}{a}{\int }_{0}^{a}\left(a-x\right)dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{b}{a}{\left[\frac{x}{2}\sqrt{{a}^{2}-{x}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{a}\right)\right]}_{0}^{a}-\frac{b}{a}{\left[ax-\frac{{x}^{2}}{2}\right]}_{0}^{a}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{b}{a}\left\{\frac{a}{2}\sqrt{{a}^{2}-{a}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\left(\frac{a}{a}\right)-\frac{0}{2}\sqrt{{a}^{2}-{0}^{2}}-\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\left(\frac{0}{a}\right)\right\}\\ \text{}\text{}\text{}\text{}-\frac{b}{a}\left\{a\left(a\right)-\frac{{a}^{2}}{2}-3\left(0\right)+\frac{{0}^{2}}{2}\right\}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{b}{a}\left\{\frac{{a}^{2}}{2}×\frac{\pi }{2}\right\}-\frac{b}{a}\left\{{a}^{2}-\frac{{a}^{2}}{2}\right\}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{b}{a}\left(\frac{\pi {a}^{2}}{4}-\frac{{a}^{2}}{2}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{b{a}^{2}}{2a}\left(\frac{\pi }{2}-1\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{ab}{4}\left(\pi -2\right)\text{\hspace{0.17em}}square\text{units}\end{array}$

Q.29 Find the area of the region enclosed by the parabola x2 = y and the line y = x + 2 and x-axis.

Ans.

The area of the region enclosed by the hyperbola and line is represented by shaded region OABCO.
The point of intersection of the parabola, x2 = y and the line, y = x + 2, is A(–1, –1).

$\begin{array}{l}\therefore Area\left(OCAO\right)={\int }_{-1}^{2}\left(x+2\right)dx-{\int }_{-1}^{2}{x}^{2}dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[\frac{{x}^{2}}{2}+2x\right]}_{-1}^{2}-{\left[\frac{{x}^{3}}{3}\right]}_{-1}^{2}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\left\{\frac{{\left(2\right)}^{2}}{2}+2\left(2\right)\right\}-\left\{\frac{{\left(-1\right)}^{2}}{2}+2\left(-1\right)\right\}\right]-\left\{\frac{{2}^{3}}{3}-\frac{{\left(-1\right)}^{3}}{3}\right\}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(2+4-\frac{1}{2}+2\right)-\left(\frac{8}{3}+\frac{1}{3}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{15}{2}-3\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{9}{2}\text{\hspace{0.17em}}square\text{units}\end{array}$

Q.30 Using the method of integration find the area bounded by the curve |x| + |y| = 1.

Ans.

The area bounded by the curve |x| + |y| = 1 can be obtained by tracing curves of the lines x + y = 1,
x – y =1, –x – y= 1 and –x + y = 1. The shaded portion ABCD is showing the required area in the graph.

The curves are intersecting the axes at the points
A (0, 1), B (1, 0), C (0, –1) and D (–1, 0).
The given curve is symmetrical about x-axis and y-axis.

$\begin{array}{l}\text{So},\text{Area ABCD}=\text{4}\left(\text{Area AOB}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4{\int }_{0}^{1}\left(1-x\right)dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4{\left[x-\frac{{x}^{2}}{2}\right]}_{0}^{1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\left(1-\frac{{1}^{2}}{2}-0+\frac{{0}^{2}}{2}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\left(\frac{1}{2}\right)=2\text{\hspace{0.17em}}square\text{units}\end{array}$

Q.31

$Find\text{}the\text{}area\text{}bounded\text{}by\text{}curves\text{\hspace{0.17em}}\left\{\left(x,y\right):y\ge {x}^{2}\text{and}y=|x|\right\}.$

Ans.

null

The required area is symmetrical about y-axis.

$\begin{array}{l}So,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{required area}=2\left\{Area\text{\hspace{0.17em}}\left(OCAO\right)-Area\text{\hspace{0.17em}}\left(OCADO\right)\right\}\\ \text{}\text{}\text{}\text{}=2\left[{\int }_{0}^{1}xdx-{\int }_{0}^{1}{x}^{2}dx\right]\\ \text{}\text{}\text{}\text{}=2{\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}-2{\left[\frac{{x}^{3}}{3}\right]}_{0}^{1}\\ \text{}\text{}\text{}\text{}=2\left[\frac{{1}^{2}}{2}-0\right]-2\left[\frac{{1}^{3}}{3}-0\right]\\ \text{}\text{}\text{}\text{}=2\left(\frac{1}{2}-\frac{1}{3}\right)\\ \text{}\text{}\text{}\text{}=2\left(\frac{1}{6}\right)=\frac{1}{3}\text{square units}\end{array}$

Q.32 Using method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3).

Ans.

The vertices of triangle ABC are A(2, 0), B(4, 5) and C(6, 3).

$\begin{array}{l}Equation\text{of line AB is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}-0=\frac{5-0}{4-2}\left(x-2\right)\\ \text{}y=\frac{5}{2}\left(x-2\right)\\ Equation\text{of line BC is}\\ \text{y}-5=\frac{3-5}{6-4}\left(x-4\right)\\ \text{}y=\frac{-2}{2}\left(x-4\right)+5\text{\hspace{0.17em}}=-x+9\\ Equation\text{of line CA is}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}-3=\frac{0-3}{2-6}\left(x-6\right)\\ \text{}y=\frac{3}{4}\left(x-6\right)+3\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3}{4}\left(x-2\right)\\ Area\left(\Delta ABC\right)=Area\left(\Delta ABLA\right)+Area\left(\Delta BLMCB\right)-Area\left(\Delta ACMA\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{2}^{4}\frac{5}{2}\left(x-2\right)dx+{\int }_{4}^{6}\left(-x+9\right)dx-{\int }_{2}^{6}\frac{3}{4}\left(x-2\right)dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{2}{\left[\frac{{x}^{2}}{2}-2x\right]}_{2}^{4}+{\left[-\frac{{x}^{2}}{2}+9x\right]}_{4}^{6}-\frac{3}{4}{\left[\frac{{x}^{2}}{2}-2x\right]}_{2}^{6}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{2}\left[\frac{{4}^{2}}{2}-2\left(4\right)-\frac{{2}^{2}}{2}+2\left(2\right)\right]+\left[-\frac{{6}^{2}}{2}+9\left(6\right)+\frac{{4}^{2}}{2}-9\left(4\right)\right]\\ \text{}\text{}\text{}\text{}\text{}-\frac{3}{4}\left[\frac{{6}^{2}}{2}-2\left(6\right)-\frac{{2}^{2}}{2}+2\left(2\right)\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{2}\left(8-8-2+4\right)+\left(-18+54+8-36\right)-\frac{3}{4}\left(18-12-2+4\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{2}\left(2\right)+8-\frac{3}{4}\left(8\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5+8-6\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=7\text{squre units}\end{array}$

Q.33 Using method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Ans.

The equations of given lines are:
2x + y = 4 … (i)
3x – 2y= 6 … (ii)
x – 3y +5 =0 … (iii)
The graphical representation of these lines is given in below. Here AL and CM are perpendicular to x-axis.

$\begin{array}{l}\mathrm{Area}\left(\mathrm{\Delta ABC}\right)=\mathrm{Area}\left(\mathrm{ALMCA}\right)-\mathrm{Area}\left(\mathrm{ALB}\right)-\mathrm{Area}\left(\mathrm{CMB}\right)\\ \mathrm{ }={\int }_{1}^{4}\frac{\mathrm{x}+5}{3}\mathrm{dx}-{\int }_{1}^{2}\left(4-2\mathrm{x}\right)\mathrm{dx}-{\int }_{2}^{4}\frac{3\mathrm{x}-6}{2}\mathrm{dx}\\ \mathrm{ }=\frac{1}{3}{\left[\frac{{\mathrm{x}}^{2}}{2}+5\mathrm{x}\right]}_{1}^{4}-{\left[4\mathrm{x}-{\mathrm{x}}^{2}\right]}_{1}^{2}-\frac{1}{2}{\left[3\frac{{\mathrm{x}}^{2}}{2}-6\mathrm{x}\right]}_{2}^{4}\\ \mathrm{ }=\frac{1}{3}\left[\frac{{4}^{2}}{2}+5\left(4\right)-\frac{{1}^{2}}{2}-5\left(1\right)\right]-\left[4\left(2\right)-{2}^{2}-4\left(1\right)+{1}^{2}\right]\\ -\frac{1}{2}\left[3\frac{{4}^{2}}{2}-6\left(4\right)-3\frac{{2}^{2}}{2}+6\left(2\right)\right]\\ \mathrm{ }=\frac{1}{3}\left(8+20-\frac{1}{2}-5\right)-\left(8-4-4+1\right)-\frac{1}{2}\left(24-24-6+12\right)\\ \mathrm{ }=\frac{1}{3}\left(\frac{45}{2}\right)-1-3\\ \mathrm{ }=\frac{15}{2}-4\\ \mathrm{ }=\frac{7}{2}\mathrm{square}\mathrm{units}\end{array}$

Q.34

$Find\text{}the\text{}area\text{}of\text{}the\text{}region\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\left(x,y\right):{y}^{2}\ge 4x,4{x}^{2}+4{y}^{2}\le 9\right\}.$

Ans.

$\begin{array}{l}\text{The intersection points of both the curves are}\text{\hspace{0.17em}}\left(\frac{1}{2},\sqrt{2}\right)\text{and}\left(\frac{1}{2},-\sqrt{2}\right).\\ The\text{required area is given by OABCO}\text{.}\\ \text{OABCO is symmetrical about x-axis}\text{.}\\ \therefore \text{Area}\text{\hspace{0.17em}}\text{OABCO}=2×Area\text{\hspace{0.17em}}OAB\\ Area\text{\hspace{0.17em}}OABO=Area\text{\hspace{0.17em}}OMAO\text{\hspace{0.17em}}+Area\text{\hspace{0.17em}}AMBA\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\frac{1}{2}}2\sqrt{x}+{\int }_{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{2}\sqrt{9-4{x}^{2}}dx\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{\frac{1}{2}}+\frac{1}{4}{\left[\frac{2x}{2}\sqrt{9-4{x}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{2x}{3}\right)\right]}_{\frac{1}{2}}^{\frac{3}{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}\left[{\left(\frac{1}{2}\right)}^{\frac{3}{2}}-0\right]+\frac{1}{4}\left[\begin{array}{l}\frac{2\left(\frac{3}{2}\right)}{2}\sqrt{9-4{\left(\frac{3}{2}\right)}^{2}}+\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{2×\frac{3}{2}}{3}\right)\\ \text{}-\frac{2×\frac{1}{2}}{2}\sqrt{9-4{\left(\frac{1}{2}\right)}^{2}}-\frac{9}{2}{\mathrm{sin}}^{-1}\left(\frac{2×\frac{1}{2}}{3}\right)\end{array}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}×\frac{1}{2\sqrt{2}}+\frac{1}{4}\left\{0+\frac{9}{2}×\frac{\pi }{2}-\frac{1}{2}×2\sqrt{2}-\frac{9}{2}×{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)\right\}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{2}}{3}+\frac{9\pi }{16}-\frac{\sqrt{2}}{4}-\frac{9}{8}{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)\end{array}$

$\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{9\pi }{16}-\frac{9}{8}{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{12}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{9\pi }{16}-\frac{9}{8}{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{12}\\ Thus,\text{the required area of shaded region is}\\ 2×\left(\frac{9\pi }{16}-\frac{9}{8}{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{12}\right)\text{\hspace{0.17em}}\text{square units}\\ \text{i}\text{.e}\text{.,}\left(\frac{9\pi }{8}-\frac{9}{4}{\mathrm{sin}}^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{6}\right)\text{square units}\text{.}\end{array}$

Q.35 Area bounded by the curve y = x3, the x-axis and the ordinates x= –2 and x = 1 is

(A) – 9 (B) – 15/4 (C) 15/4 (D) 17/4

Ans.

The required area is shaded in the figure.

$\begin{array}{l}\mathrm{Re}quired\text{area}=|{\int }_{-2}^{0}{x}^{3}dx|+{\int }_{0}^{1}{x}^{3}dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|{\left[\frac{{x}^{4}}{4}\right]}_{-2}^{0}|+{\left[\frac{{x}^{4}}{4}\right]}_{0}^{1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\frac{1}{4}\left(0-16\right)|+\frac{1}{4}\left(1-0\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{16}{4}+\frac{1}{4}\\ \text{}\text{}\text{}=\frac{17}{4}\\ Thus,\text{correct option is D}\text{.}\end{array}$

Q.36 The area bounded by the curve y = x |x|, x-axis and the ordinates x = –1 and x = 1 is given by

(A) 0 (B) 1/3 (C) 2/3 (D) 4/3

Ans.

$\begin{array}{l}We\text{have,}\\ y=x|x|=\left\{\begin{array}{l}-{x}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 0\end{array}\end{array}$

$\begin{array}{l}\text{Intersection points of curve x}|x|\text{and lines x}=1\text{}and\text{}x=-\text{1 are}\\ \left(1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\right)\text{and}\left(-1,-1\right)\text{respectively}\text{.}\\ \therefore \text{Required area}={\int }_{-1}^{1}x|x|dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|{\int }_{-1}^{0}-{x}^{2}dx|+{\int }_{0}^{1}{x}^{2}dx\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|{\left[-\frac{{x}^{3}}{3}\right]}_{-1}^{0}|+{\left[\frac{{x}^{3}}{3}\right]}_{0}^{1}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\left(0+\frac{1}{3}\right)|+\left(\frac{1}{3}-0\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3}+\frac{1}{3}\text{\hspace{0.17em}}=\frac{2}{3}\text{square units}\\ \text{Thus, the correct option is C}\text{.}\end{array}$

Q.37

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{circle}{\mathrm{x}}^{\mathrm{2}}+{\mathrm{y}}^{\mathrm{2}}\mathrm{ }=16\mathrm{exterior}\mathrm{to} \mathrm{the}\mathrm{parabola}\\ {\mathrm{y}}^{\mathrm{2}}\mathrm{ }=6\mathrm{x}\mathrm{is}\\ \left(\mathrm{A}\right)\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{4}\mathrm{\pi }–\sqrt{\mathrm{3}}\right)\left(\mathrm{B}\right)\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{4}\mathrm{\pi } +\mathrm{ }\sqrt{\mathrm{3}}\right)\\ \left(\mathrm{C}\right) \frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{8}\mathrm{\pi }–\sqrt{\mathrm{3}}\right)\left(\mathrm{D}\right)\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{8}\mathrm{\pi } +\mathrm{ }\sqrt{\mathrm{3}}\right)\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\mathrm{intersection}\mathrm{points}\mathrm{of}\mathrm{parabola}\mathrm{and}\mathrm{circle}\mathrm{are}\mathrm{A}\mathrm{ }\left(2,2\sqrt{3}\right)\mathrm{ }\mathrm{and}\mathrm{ }\\ \mathrm{B}\mathrm{ }\left(2,-2\sqrt{3}\right).\mathrm{ }\mathrm{Area}\mathrm{of}\mathrm{circle}\mathrm{which}\mathrm{is}\mathrm{exterior}\mathrm{to}\mathrm{parabola}\mathrm{is}\mathrm{shown}\\ \mathrm{by}\mathrm{shaded}\mathrm{area}\mathrm{.}\end{array}$

$\begin{array}{l}\therefore Unshaded\text{\hspace{0.17em}}Area\text{bounded by circle and parabola}\\ \text{}\text{}\text{}\text{}\text{}=2\left({\int }_{0}^{2}\sqrt{6x}dx+{\int }_{2}^{4}\sqrt{16-{x}^{2}}dx\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\sqrt{6}{\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{0}^{2}+{\left[\frac{x}{2}\sqrt{16-{x}^{2}}+\frac{16}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{4}\right)\right]}_{2}^{4}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\frac{2\sqrt{6}}{3}\left[{2}^{\frac{3}{2}}-0\right]+{\left[\begin{array}{l}0+8{\mathrm{sin}}^{-1}\left(1\right)-\sqrt{12}\\ \text{}-8{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)\end{array}\right]}_{2}^{4}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\frac{8\sqrt{3}}{3}+8×\frac{\pi }{2}-2\sqrt{3}-8×\frac{\pi }{6}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(\frac{2\sqrt{3}}{3}+\frac{8\pi }{3}\right)=\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}\\ Area\text{of shaded region}=Area\text{of circle}-\left(\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\pi {\left(4\right)}^{2}-\left(\frac{4\sqrt{3}}{3}+\frac{16\pi }{3}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=16\pi -\frac{16\pi }{3}-\frac{4\sqrt{3}}{3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{32\pi }{3}-\frac{4\sqrt{3}}{3}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{4}{3}\left(8\pi -\sqrt{3}\right)\text{square units}\\ \text{Thus, correct option is C}\text{.}\end{array}$

Q.38

$\begin{array}{l}\text{The area bounded by the y-axis, y = cosx and y = sinx}\\ \text{when 0}\le \text{x}\le \frac{\pi }{\text{2}}\text{is}\\ \text{}\left(\text{A}\right)\text{2}\left(\sqrt{\text{2}}-\text{1}\right)\text{}\left(\text{B}\right)\text{}\sqrt{\text{2}}-\text{1}\text{}\left(\text{C}\right)\sqrt{\text{2}}\text{+1}\text{}\left(\text{D}\right)\sqrt{\text{2}}\end{array}$

Ans.

$\begin{array}{l}Given\text{curves are y = sinx and y = cosx}\\ \text{Intersection point of both given curves is B}\left(\frac{\pi }{4},\text{\hspace{0.17em}}\frac{1}{\sqrt{2}}\right).\\ \therefore Area\text{of shaded region}=Area\left(AMBA\right)+Area\left(OMBO\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\frac{1}{\sqrt{2}}}x\text{\hspace{0.17em}}dy+{\int }_{\frac{1}{\sqrt{2}}}^{1}x\text{\hspace{0.17em}}dy\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\frac{1}{\sqrt{2}}}{\mathrm{sin}}^{-1}y\text{\hspace{0.17em}}dy+{\int }_{\frac{1}{\sqrt{2}}}^{1}{\mathrm{cos}}^{-1}y\text{\hspace{0.17em}}dy\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\int }_{0}^{\frac{1}{\sqrt{2}}}{\mathrm{sin}}^{-1}y.1\text{\hspace{0.17em}}dy+{\int }_{\frac{1}{\sqrt{2}}}^{1}{\mathrm{cos}}^{-1}y.1\text{\hspace{0.17em}}dy\\ Integrating\text{by parts, we get}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[y{\mathrm{sin}}^{-1}y+\sqrt{1-{y}^{2}}\right]}_{0}^{\frac{1}{\sqrt{2}}}+{\left[y{\mathrm{cos}}^{-1}y-\sqrt{1-{y}^{2}}\right]}_{\frac{1}{\sqrt{2}}}^{1}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\frac{1}{\sqrt{2}}{\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^{2}}-0-1\right]\\ \text{}\text{}\text{}\text{}+\left[1{\mathrm{cos}}^{-1}\left(1\right)-\sqrt{1-{\left(1\right)}^{2}}-\frac{1}{\sqrt{2}}{\mathrm{cos}}^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^{2}}\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\frac{1}{\sqrt{2}}\frac{\pi }{4}+\frac{1}{\sqrt{2}}-1\right]+\left[0-0-\frac{1}{\sqrt{2}}\frac{\pi }{4}+\frac{1}{\sqrt{2}}\right]\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{\sqrt{2}}\frac{\pi }{4}+\frac{1}{\sqrt{2}}-1-\frac{1}{\sqrt{2}}\frac{\pi }{4}+\frac{1}{\sqrt{2}}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{2}{\sqrt{2}}-1=\sqrt{2}-1\\ Thus,\text{the correct option is B}\text{.}\end{array}$