# NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex 8.1) Exercise 8.1

The National Council of Educational Research and Training (NCERT) is a self-governing body established by the Indian government in 1961 set up a body to establish ground rules for education. To assist and provide advice to the Central and State Governments on policies and programmes aimed at improving the quality of schooling. The primary goals of NCERT and the units that make up the organisation are to conduct, promote, and coordinate research in fields related to school education; create and publish model Textbooks, Supplemental Materials, Newsletters, Journals, and Educational Kits; and develop Multimedia Digital Materials, among other things.

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## NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals (Ex. 8.1) Exercise 8.1

The most important thing for students to do in subjects like mathematics is to practise as much as they can.NCERT Solutions for Class 12 Maths, Chapter 8, Exercise 8.1, can be used by students for practise and revision.

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### Access NCERT Solutions for Class 12 Maths Chapter 08-Application of Integrals

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### Access NCERT Solutions for Class 12 Maths Chapter 08-Application of Integrals

Exercise 8.1 Class 12th is the first exercise of Chapter 8. Class 12 Chapter 8 is based on the Application of Integrals. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 was created keeping in mind the latest syllabus issued by the CBSE.

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### Topics Covered in Class 12 Maths Chapter 8 Exercise 8.1

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 for Mathematics for grade 12 focus on applications of integrals. Applications of Integrals for Class 12 include determining the area of the functions and curves, which is nothing more than finding the sum of the integrations that were taught to the students of Class 12 in Chapter 7 of Mathematics.

The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 can be very useful for the students who wish to practice the questions that are given in the NCERT books of Mathematics for Class 12 chapter 8.

Chapter 8 includes the following topics:

The Region Enclosed by a Curve, the Region Enclosed by Two Curves, The Area Under a Curve, and the Area Under a Curve and a Line.

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### How to Find the Area Under a Simple Curve?

The topics that are covered in Class 12 Chapter 8 Mathematics are some of the important topics that are also covered in the syllabus for competitive exams like the JEE Main/NEET etcetera. One such topic is – Areas and Simple Curves. It is the method of calculating the area under the Curve. The students get to learn how to find the area of different shapes of objects like Trapezium, Triangles, Rectangles etcetera.

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### NCERT Solutions for Class 12 Maths Chapter 8 Exercises

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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

$\begin{array}{l}\mathrm{Find the area of the region in the first} \mathrm{quadrant enclosed}\\ \mathrm{by x}–\mathrm{axis},\mathrm{line x}=\sqrt{3}{\mathrm{yandthecirclex}}^{2}+{\mathrm{y}}^{2}=4.\end{array}$

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}$

## 1. Is the NCERT book for Mathematics enough for CBSE Class 12 Mathematics exam preparation?

The question papers for Board examinations are based on the NCERT syllabus.

So the students are advised to first complete the NCERT book when preparing for their board examinations. But when it comes to subjects like Mathematics, the more the students practice, the better it is for their performance. Once the students are done with their NCERT books, they are advised to look for other helpful tools based on the NCERT book that can help them revise and practice. The Extramarks’ website has all such tools easily available for the students to access.

## 2. Is the syllabus for Class 12 Mathematics very difficult for the students to score full marks?

For subjects like Mathematics, when compared to language subjects, it becomes easier and more plausible to score full marks. The students should remember to invest a lot of time in practise and revision when preparing for subjects like Mathematics. With the use of the right tools for preparation, studentscan achieve excellent marks in their board examinations. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 can be a great tool for revision and practice.

## 3. Is Class 12 Mathematics useful for students after school?

The study material of Class 12 Mathematics is prepared in a way that is useful  to the students even after they finish their schooling. Class 12 students are expected to appear for many other competitive examinations once they are done with their board examinations.

These competitive exams are taken for the students to decide how they wish to continue their higher education, in which field, and at which college. The NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1 can also be very helpful in preparing for these examinations. The Mathematics that is learned by Class 12 students for their board examinations can be of a lot of use when they are preparing for these competitive examinations. A lot of times, what is taught in college education is an extension of what the students learned in their 12th grade year.