# CBSE Class 12 Maths Revision Notes Chapter 8

## NCERT Class 12  Mathematics Chapter 8 Notes

Mathematics is an essential subject that forms the basis of all other subjects. It encourages logical reasoning and is necessary for business, finance and personal decision making. In this subject, strong basic knowledge is  required to understand complex subjects and solve problems based on them.. The majority of students face issues and find it difficult to tackle problems in mathematics. To easily understand the concepts, methods and gain knowledge of the higher-level topics, students must  learn from Class 12 Mathematics Chapter 8 notes and get an understanding of all the basic concepts.

The Class 12 Chapter 8 Mathematics notes- Application of Integrals is a continuation of Chapter 7. Integration is an important concept  in higher mathematics. This Chapter deals with the calculation of complex areas under curves, parabolas, ellipses, and intercepts. Students also learn about various integration techniques and formulas. With the help of the Class 12 Mathematics Chapter 8 notes, students can strengthen their foundation and solve problems based on the real world.

Extramarks make Mathematics interesting and fun with coloured illustrations and detailed information. The Class 12 Mathematics Chapter 8 notes help students to practice unlimited questions and attain high scores in the examination. Students can also access different practice tests, mock tests, reference books, etc.,  from the Extramarks web portal.

## NCERT Class 12 Mathematics Chapter 8: Key Notes

The main topics covered in the Class 12 Mathematics Chapter 8 notes are:

• Introduction
• Definite integrals
• Newton Leibnitz’s Theorem
• Area under Simple Curves
• Area between Two Curves.

### Introduction:

The Chapter begins by recalling the concepts of finding areas bounded by the curve and definite integrals. It also introduces the various applications of integrals in this Chapter and topics like areas under simple curves. Between lines,  and curves, parabola, and ellipses are also taught. Calculation of the average value of a function with the help of integration is included in the Chapter.

Real-life problems such as a record of rainfall in a day expressed in the form of a curve with specified limit x to limit y are included in the Class 12 Mathematics Chapter 8 notes.

### Definite Integrals:

Let F(x) be the antiderivative of function f(x), then the definite integral of f(x) from a to b is given as F(b) – F(a), such that variable x has any two independent values a and b. It is denoted as abf(x) dx.

Therefore, we can say that  abf(x) dx= F(b) – F(a). the values a and b are called the limits of integration.

Properties:

1. abf(x) dx = –baf(x) dx
2. abf(x) dx = baf(y) dy
3. abf(x) dx = acf(x) dx + cbf(x) dx, where a< c <b
4. 0af(x) dx = abf(a-x) dx
5. abf(x) dx =abf(a+b-x) dx
6. 0af(x)f(x)+f(a-x) dx = a2
7. abf(x)f(x)+f(a+b-x) dx = b-a2
8. 02af(x) dx = 0af(x) dx + 0af(2a-x) dx

If f(x) is a periodic function, i.e. f(a+x) = f(x) then,

1.  0naf(x) dx=  n0af(x) dx
2.  0naf(x) dx= (n-1)  0af(x) dx
3.  0b+naf(x) dx0bf(x) dx
4. If f(x) 0 on [a, b], then  abf(x) dx 0
5. If f(x) g(x) on [a, b], then   abf(x) dx   abg(x) dx
6. abf(x) dx abf(x) dx
7. abf(x) dx = f(c) (b – a), for a < c < b

#### Newton Leibnitz’s Theorem:

Consider two differentiable functions g(x) and h(x) for x [a, b], the function f is continuous in interval [a, b] then

ddxg(x)h(x)f(x) dx =ddxh(x) . f(h(x)) – ddxg(x) . f(g(x))

### Definite Integral as a Limit of Sum:

The function f(x) is continuous on the interval [a, b] divided into n parts, then

abf(x) dx= nr=0n=1(b-a)n f(a+ (b-a) rn

### Reduction Formulae in Definite Integrals

1. If In= o2sinnx dx then In=(n-1n) In-2

NOTE: In= o2sinnx dx  = o2cosnx dx

1. If In= o4tannx dx then In+ In-2= 1n-1
2. If In= o2sinmx. cosnx dx then Im,n=(m-1m+n) Im-2, n

### The area under the curves:

1. The total area A of the region bounded between the x-axis with co-ordinates x = a, x = b and the curve y = f (x)  is given as abdA = aby dx =abf(x) dx.

If f(x)>0, ∀x∈[a,c) and f(x)<0 ∀x∈(c,b], then

Area = acf(x) dx+ cbf(x) dx = acf(x) dxcbf(x) dx  ∀ a < c < b

2. The area of the given region bounded between the curve x = g (y), y-axis and the lines y = c,

y = d is given as cddA = cdx dy =cdg(y) dx. NOTE: If the curve is below the x-axis, then f (x) < 0 from x = a to x = b. In this case, the area will be negative. But since only the numerical value is taken into consideration, we take the absolute value of the area, which is given by abf(x) dx

### The area of the region bounded by a curve and a line:

The area of the region is bounded by a line and a circle, parabola, or an ellipse in their standard forms. Vertical stripes or horizontal stripes are used to calculate the area of the region.

For, e.g. Consider the figure given below. We have to find the area bounded by the ellipse in its standard form and the ordinates x = 0 and x = ae, where b2= a2(1- e2) and e < 1. The area of the highlighted region is enclosed by the lines x = 0 and x = ae and the eclipse. Therefore using the formula for area, we will integrate and find the solution. ### The area of the region between two curves: Here total area = ab[f(x)-g(x)]dx i.e.,

Area A= [area bounded by the curve y = f (x), x-axis and lines x = a, x = b] – [area bounded by the curve

y = g (x), x-axis and lines x = a, x = b]

A = abf(x)dx –abg(x)dxab[f(x)-g(x)]dx where f(x) > g(x) in the interval [a, b] In this case, the total area A= Area of the ACBDA region+ Area of the BPRQB region.

Area = ac[f(x)-g(x)]dx + cb[g(x)-f(x)]dx, where a < c < b and f (x) ≥ g (x) in interval [a, c] and f (x) ≤ g (x) in the interval [c, b].

## Curve Tracing:

To locate the area of a region, it is necessary to draw a rough sketch. Consider the curve f(x,y) = 0. To find the area of the curve, follow the steps given below:

### Step 1: Symmetry

• The curve is said to be symmetric about the x-axis if all the powers of y in the equation are even.
• The curve will be symmetric about the y-axis if all the powers of x in the equation are even.
• The curve is symmetric about line y = x if the given equation remains unchanged on interchanging the value of x and y.
• The curve will be symmetrical in opposite quadrants if the given equation remains unchanged when x and y values are replaced by -x and -y.

### Step 2: Origin

• If the constant term is absent in the given equation, then we can say that the curve passes through the origin (0,0).
• Then calculate the tangents at the point (0, 0) by equating the terms having the lowest degree in the given equation to zero.

### Step 3: Intersection with Co-ordinates Axes

• Find values of x by substituting y=0 to estimate the intersecting points of the curve with an x-axis
• Find values of y by substituting x=0 to estimate the intersecting points of the curve with the y-axis

### Step 4: Asymptotes

• Compare the coefficient of the highest power of variable y in the given algebraic equation to zero to find out the vertical asymptotes.
• Compare the coefficient of the highest power of variable x in the given algebraic equation to zero to find out the horizontal asymptotes.

### Step 5: Region

• Solve the given algebraic equation for x in terms of y or vice versa to determine the regions in which the curve doesn’t lie.

### Step 6: Critical Points

• Differentiate the value of y with respect to x and find out which values of x satisfy ddxy = 0

## Chapter 8 Mathematics Class 12 Notes: Exercises & Answer Solutions

The Class 12 Mathematics Chapter 8 notes ensure detailed and apt information of all the concepts for students to get a clear understanding. Students learn to calculate the area of different regions bounded by curves, lines, parabolas and ellipses. With the help of the Class 12 Mathematics Chapter 8 notes, students can get all important definitions, formulas, properties, and theorems in one place to enable quick revision to clear their doubts and provide them with a solid foundation..

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• It is prepared by subject matter experts at Extramarks
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Q.1 Find the area of the region bounded by y2 = x, x =1, x = 4 and the x-axis.

Ans $\begin{array}{l}{\mathrm{y}}^{2}=\mathrm{x} \mathrm{trepresents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\left(0,\mathrm{ }0\right).\\ \mathrm{x}\mathrm{ }=\mathrm{ }1\mathrm{and}\mathrm{x}= 4\mathrm{represent}\mathrm{the}\mathrm{straight}\mathrm{lines}\mathrm{parallel}\mathrm{to}\mathrm{y}–\mathrm{axis}.\\ \mathrm{The}\mathrm{required}\mathrm{area}=\underset{1}{\overset{4}{\int }}\sqrt{\mathrm{x}}\mathrm{dx}\\ = {\left[\frac{2{\mathrm{x}}^{\frac{3}{2}}}{3}\right]}_{1}^{4}\\ =\mathrm{ }\frac{2}{3}\left[4\sqrt{4}-1\sqrt{1}\right]\\ =\mathrm{ }\frac{14}{3}\mathrm{sq}.\mathrm{units}\end{array}$

Q.2 Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = – 1 and x = 1.

Ans $\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{line}\mathrm{meets}\mathrm{x}–\mathrm{axis}\mathrm{at}\mathrm{x}= –\frac{2}{3}\mathrm{and}\mathrm{its}\mathrm{graph}\mathrm{lies}\mathrm{below}\mathrm{x}–\mathrm{axis}\mathrm{for}\mathrm{x}\in \mathrm{}\left(-1,\mathrm{ }\frac{2}{3}\right)\\ \mathrm{and}\mathrm{above}\mathrm{x}–\mathrm{axis}\mathrm{for}\mathrm{x}\in \mathrm{}\left(-\frac{2}{3},\mathrm{ }1\right).\\ \mathrm{The}\mathrm{required}\mathrm{area}=\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{region}\mathrm{ACBA}+\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{region}\mathrm{ADEA}\\ = \left|\underset{-1}{\overset{-\frac{2}{3}}{\int }}\left(3\mathrm{x}+2\right)\mathrm{dx}\right|+\underset{-\frac{2}{3}}{\overset{1}{\int }}\left(3\mathrm{x}+2\right)\mathrm{dx}\\ = \left|{\left[\frac{3{\mathrm{x}}^{2}}{2}+2\mathrm{x}\right]}_{01}^{-\frac{2}{3}}\right|+{\left[\frac{3{\mathrm{x}}^{2}}{2}+2\mathrm{x}\right]}_{-\frac{2}{3}}^{1}\\ =\mathrm{ }\frac{1}{6}+\frac{25}{6}\\ =\mathrm{ }\frac{13}{3}\mathrm{sq}.\mathrm{units}\end{array}$

Q.3 Using integration, find the area bounded by the curves x2 + y2 = 1 and y2 = (x +1).

Ans $\begin{array}{l}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=1 \mathrm{represents}\mathrm{circle}\mathrm{with}\mathrm{centre}\left(0,\mathrm{ }0\right) \mathrm{and}\mathrm{radius}= 1.\\ {\mathrm{y}}^{2}\mathrm{ }=\mathrm{ }\left(\mathrm{x}+1\right) \mathrm{represents}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\left(-1,\mathrm{ }0\right).\\ \mathrm{From}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+1=\mathrm{ }1\mathrm{and}{\mathrm{y}}^{2}=\left(\mathrm{x}+1\right), \mathrm{we}\mathrm{get}\\ {\mathrm{x}}^{2}+\mathrm{x}+1=\mathrm{ }1\\ \mathrm{or} {\mathrm{x}}^{2}+\mathrm{x}\mathrm{ }=0\\ \mathrm{or} {\mathrm{x}}^{2}\left(\mathrm{x}+1\right)=\mathrm{ }0\\ \mathrm{o} \mathrm{x}\mathrm{ }= 0,\mathrm{ }-1\\ \mathrm{The}\mathrm{required}\mathrm{area}= 2\mathrm{x}\left[\underset{-1}{\overset{0}{\int }}\sqrt{1-{\mathrm{x}}^{2}}\mathrm{dx}-\underset{-1}{\overset{0}{\int }}\sqrt{\mathrm{x}-1}\mathrm{dx}\right]\\ \mathrm{The}\mathrm{required}\mathrm{area}= 2\mathrm{x} {\left[\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^{2}}+\frac{1}{2}{\mathrm{sin}}^{-1}\mathrm{x}-\frac{2}{3}{\left(\mathrm{x}+1\right)}^{\frac{3}{2}}\right]}_{-1}^{0}\\ \mathrm{The}\mathrm{required}\mathrm{area}= 2\mathrm{x} \left[-\frac{2}{3}-\left(-\frac{\mathrm{\pi }}{4}\right)\right]\\ \mathrm{The}\mathrm{required}\mathrm{area}=\frac{\mathrm{\pi }}{2}-\frac{4}{3}\mathrm{sq}.\mathrm{units}\end{array}$

Q.4 Find the area bounded by curves x2 + y2 = 1 and (x – 1)2 + y2 = 1.

Ans $\begin{array}{l}\mathrm{Equations}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{circles}\mathrm{are}\mathrm{as}\mathrm{follows}:\\ {\mathrm{x}}^{2}+{\mathrm{y}}^{2}=\mathrm{ }1 .\dots .\mathrm{ }\left(1\right)\\ \mathrm{and}{\left(\mathrm{x}-1\right)}^{2}+{\mathrm{y}}^{2}=1\mathrm{ }.\dots .\left(2\right)\\ \mathrm{Equation}\left(1\right) \mathrm{is}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centra}=\left(0,\mathrm{ }0\right)\mathrm{and}\mathrm{radius}= 1.\\ \mathrm{Equation}\left(2\right)\mathrm{is}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centra}=\left(1,\mathrm{ }0\right)\mathrm{and}\mathrm{radius}= 1.\\ {\mathrm{x}}^{\mathrm{ }2}+1-{\left(\mathrm{x}-1\right)}^{2}=1\\ \mathrm{or} {\mathrm{x}}^{2}-{\mathrm{x}}^{2}-1+2\mathrm{x}\mathrm{ }=0\\ \mathrm{or} \mathrm{x}=\mathrm{ }\frac{1}{2}\\ \mathrm{Required}\mathrm{area}= 2\mathrm{x}\left[\underset{0}{\overset{\frac{1}{2}}{\int }}\sqrt{1-{\left(\mathrm{x}-1\right)}^{2}}\mathrm{dx}+\underset{0}{\overset{\frac{1}{2}}{\int }}\sqrt{1-\left(1-{\mathrm{x}}^{2}\right)}\mathrm{dx}\right]\\ =\mathrm{ }2\mathrm{x}\left\{{\left[\frac{\mathrm{x}-1}{2}\sqrt{1-{\left(\mathrm{x}-1\right)}^{2}+\frac{1}{2}}{\mathrm{sin}}^{-1}\left(\mathrm{x}-1\right)\right]}_{0}^{\frac{1}{2}}+{\left[\frac{\mathrm{x}}{2}\sqrt{1-{\mathrm{x}}^{2}+\frac{1}{2}}{\mathrm{sin}}^{-1}\right]}_{\frac{1}{2}}^{1}\mathrm{ }\right\}\mathrm{ }\\ =\mathrm{ }2\mathrm{x}\left\{{\left[\left(-\frac{\sqrt{3}}{8}-\frac{\mathrm{\pi }}{12}\right)+\frac{\mathrm{\pi }}{4}\right]}_{}^{}+{\left[\frac{\mathrm{\pi }}{4}-\left(\frac{\sqrt{3}}{8}+\frac{\mathrm{\pi }}{12}\right)\right]}_{}^{}\right\}\\ =\mathrm{ }2\mathrm{x}\left\{-\frac{\sqrt{3}}{8}-\frac{\mathrm{\pi }}{12}+\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}+\frac{\sqrt{3}}{8}-\frac{\mathrm{\pi }}{12}\right\}\\ =\mathrm{ }2\mathrm{x}\left\{-\frac{\sqrt{3}}{4}-\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{2}\right\}\\ =\mathrm{ }2\mathrm{x}\left\{-\frac{\sqrt{3}}{4}+\frac{2\mathrm{\pi }}{6}\right\}\\ =\mathrm{ }\left\{-\frac{2\mathrm{\pi }}{3}-\frac{\sqrt{3}}{2}\right\}\mathrm{sq}.\mathrm{units}.\end{array}$

Q.5

$\text{Using integration, find the area of ΔABC where A is}\left(2,3\right),\mathrm{B}\mathrm{is}\left(4,\mathrm{ }7\right)\mathrm{and}\mathrm{C}\mathrm{is}\left(6,\mathrm{ }2\right).$

Ans $\begin{array}{l}\mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{side}\mathrm{AB}\mathrm{is}\\ \mathrm{y}-3 =\frac{7-3}{4-2}\mathrm{ }\left(\mathrm{x}-2\right)\\ \mathrm{or} \mathrm{y}=2\mathrm{x}-1 .\dots ..\left(\mathrm{i}\right)\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{side}\mathrm{BC}\mathrm{is}\\ \mathrm{y}-7 = \frac{2-7}{6-4}\left(\mathrm{x}-4\right)\\ \mathrm{or}\mathrm{ }\mathrm{y}=-\frac{5}{2}\mathrm{x}+17 .\dots \left(\mathrm{ii}\right)\\ \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{side}\mathrm{AC}\mathrm{is}\\ \mathrm{y}-3 =\frac{2-3}{6-2}\left(\mathrm{x}-2\right)\\ \mathrm{or} \mathrm{y}\mathrm{ }=\mathrm{ }\frac{-1}{4}\mathrm{x}+\frac{7}{2} .\dots .\mathrm{ }\left(\mathrm{iii}\right)\\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\underset{2}{\overset{4}{\int }}{\mathrm{y}}_{\mathrm{AB}}\mathrm{dx}+\underset{4}{\overset{6}{\int }}{\mathrm{y}}_{\mathrm{BC}}\mathrm{dx}-\underset{2}{\overset{6}{\int }}{\mathrm{y}}_{\mathrm{AC}}\mathrm{dx}\\ =\underset{2}{\overset{4}{\int }}\left(2\mathrm{x}-1\right)\mathrm{dx}+\underset{4}{\overset{6}{\int }}\left(-\frac{5}{2}\mathrm{x}+17\right)\mathrm{dx}-\underset{2}{\overset{6}{\int }}\left(\frac{-1}{4}\mathrm{x}+\frac{7}{2}\right)\mathrm{dx}\\ =\mathrm{ }{\left[{\mathrm{x}}^{2}-\mathrm{x}\right]}_{2}^{4}+{\left[-\frac{5}{2}{\mathrm{x}}^{2}+17\mathrm{x}\right]}_{4}^{6}-{\left[-\frac{{\mathrm{x}}^{2}}{8}+\frac{7}{2}\mathrm{x}\right]}_{4}^{6}\\ =\mathrm{ }\left[12-2\right]+\left[\left(-45+102\right)-\left(20+68\right)\right]-\left[\left(-\frac{9}{2}+21\right)-\left(-\frac{1}{2}+7\right)\right]\\ =\mathrm{ }\left[10\right]+\left[57-48\right]-\left[-4+14\right]\\ =\mathrm{ }10+9-10\\ =\mathrm{ }9\mathrm{sq}.\mathrm{units}\end{array}$

Q.6

$\text{Find the area of the region}\left\{\mathrm{x},\mathrm{y}: 0\le \mathrm{y}\le {\mathrm{x}}^{2}+1,\mathrm{ }0\mathrm{ }\le \mathrm{y}\le \mathrm{y}\le \mathrm{x}+1,\mathrm{ }0\mathrm{ }\le \mathrm{x}\le 2\right\}.$

Ans $\begin{array}{l}\mathrm{y}={\mathrm{x}}^{2}+1\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}.\\ \mathrm{y}=\mathrm{x}+1\mathrm{represents}\mathrm{a}\mathrm{straight}\mathrm{line}.\\ \mathrm{From}\mathrm{y}={\mathrm{x}}^{2}+1\mathrm{and}\mathrm{y}=\mathrm{x}+1,\mathrm{we}\mathrm{get}\\ \mathrm{x}+1={\mathrm{x}}^{2}+1\\ \mathrm{or} \mathrm{x}-{\mathrm{x}}^{2}\mathrm{ }=0\\ \mathrm{or}\mathrm{x}\left(1-\mathrm{x}\right)=0\\ \mathrm{or} \mathrm{x}\mathrm{ }=\mathrm{ }0,\mathrm{ }1\\ \mathrm{The}\mathrm{required}\mathrm{area}=\underset{0}{\overset{1}{\int }}\left({\mathrm{x}}^{2}+1\right)\mathrm{dx}+\underset{0}{\overset{1}{\int }}\left(\mathrm{x}+1\right)\mathrm{dx}\\ = {\left[\frac{{\mathrm{x}}^{3}}{3}+\mathrm{x}\right]}_{0}^{1}+{\left[\frac{{\left(\mathrm{x}+1\right)}^{2}}{2}\right]}_{1}^{2}\\ =\mathrm{ }\left[\frac{1}{3}+1\right]+\left[\frac{9}{2}-\frac{4}{2}\right]\\ =\mathrm{ }\frac{4}{5}+\frac{5}{2}\\ =\mathrm{ }\frac{8+15}{6}\\ =\mathrm{ }\frac{23}{6}\mathrm{sq}.\mathrm{units}\\ \mathrm{Find}\mathrm{}\mathrm{the}\mathrm{}\mathrm{area}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{region}\mathrm{}\left\{\mathrm{x},\mathrm{}\mathrm{y}:\mathrm{}0\mathrm{y}{\mathrm{x}}^{2}+\mathrm{}1,\mathrm{}0\mathrm{y}\mathrm{y}\mathrm{x}\mathrm{}+1,\mathrm{}0\mathrm{x}2\right\}.\\ \mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{region}\left\{\mathrm{x},\mathrm{y}: 0\le \mathrm{y}\le {\mathrm{x}}^{2}+1,\mathrm{ }0\mathrm{ }\le \mathrm{y}\le \mathrm{y}\le \mathrm{x}+1,\mathrm{ }0\mathrm{ }\le \mathrm{x}\le 2\right\}.\end{array}$

Q.7 Find the area of the region bounded by y2 = x and x2 = y.

Ans $\begin{array}{l}{\mathrm{y}}^{2}=\mathrm{x}\mathrm{and}{\mathrm{x}}^{2}=\mathrm{y}\mathrm{represents}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\left(0,\mathrm{ }0\right).\\ \mathrm{From}{\mathrm{y}}^{2}=\mathrm{x}\mathrm{and}{\mathrm{x}}^{2}=\mathrm{y},\mathrm{we}\mathrm{get}\\ {\mathrm{x}}^{4}=\mathrm{x}\\ \mathrm{or} {\mathrm{x}}^{4}-\mathrm{x}\mathrm{ }=0\\ \mathrm{or}\mathrm{x}\left({\mathrm{x}}^{3}-1\right)=0\\ \mathrm{or} \mathrm{x}\mathrm{ }=\mathrm{ }0,\mathrm{ }1\\ \mathrm{The}\mathrm{required}\mathrm{area}=\underset{0}{\overset{1}{\int }}\mathrm{x}\mathrm{dx}-\underset{0}{\overset{1}{\int }}{\mathrm{x}}^{2}\mathrm{dx}\\ = {\left[\frac{2{\mathrm{x}}^{2}}{2}-\frac{{\mathrm{x}}^{3}}{3}\right]}_{0}^{1}\\ =\mathrm{ }\frac{2}{3}-\frac{1}{3}\\ =\mathrm{ }\frac{1}{3}\mathrm{sq}.\mathrm{units}\end{array}$

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

Ans $\begin{array}{l}\mathrm{From}\mathrm{y}=\left|\mathrm{x}\right| \mathrm{and}{\mathrm{x}}^{2}=\mathrm{y},\mathrm{we}\mathrm{get}\\ \mathrm{x}= 1, -1\\ \mathrm{The}\mathrm{required}\mathrm{area}= 2\mathrm{x}\left[\underset{0}{\overset{1}{\int }}{\mathrm{y}}_{\mathrm{L}}\mathrm{dx}-\underset{0}{\overset{1}{\int }}{\mathrm{y}}_{\mathrm{p}}\mathrm{dx}\right]\\ =\mathrm{ }2\mathrm{x}\left[\underset{0}{\overset{1}{\int }}\mathrm{xdx}-\underset{0}{\overset{1}{\int }}{\mathrm{x}}^{2}\mathrm{dx}\right]\\ =\mathrm{ }2\mathrm{x}\mathrm{ }\frac{{\mathrm{x}}^{2}}{2}-{\frac{{\mathrm{x}}^{3}}{3}}_{0}^{1}\\ =\mathrm{ }2\mathrm{x}\left[\frac{1}{2}-\frac{1}{3}\right]\\ =\mathrm{ }2\mathrm{x}\left[\frac{1}{6}\right]\\ =\mathrm{ }\frac{1}{3}\mathrm{sq}.\mathrm{units}\end{array}$

Q.9 Find the area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x.

Ans $\begin{array}{l}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=16\mathrm{represents}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{at}\left(0,\mathrm{ }0\right) \mathrm{and}\mathrm{radius}= 4\mathrm{units}.\\ {\mathrm{y}}^{2}=\mathrm{ }6\mathrm{x}\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}=\left(0,\mathrm{ }0\right).\\ \mathrm{From}\mathrm{the}\mathrm{above}\mathrm{equations}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=\mathrm{ }16\mathrm{and}{\mathrm{y}}^{2}\mathrm{ }=\mathrm{ }6\mathrm{x}, \mathrm{we}\mathrm{get}\\ {\mathrm{x}}^{2}+6\mathrm{x}-16=0\\ \mathrm{or} {\mathrm{x}}^{2}+8\mathrm{x}-2\mathrm{x}-16=0\\ \mathrm{or} \left(\mathrm{x}+8\right)\left(\mathrm{x}+2\right)\mathrm{ }=\mathrm{ }0\\ ⇒ \mathrm{x}=\mathrm{ }2,\mathrm{ }\mathrm{or}\mathrm{ }-8\mathrm{}\left(\mathrm{neglected}\right)\\ \mathrm{Thus}\mathrm{the}\mathrm{required}\mathrm{area}= 16\mathrm{\pi }–2×\mathrm{}\left[\underset{0}{\overset{2}{\int }}\sqrt{6\mathrm{xdx}}+\underset{2}{\overset{4}{\int }}\sqrt{16-{\mathrm{x}}^{2}}\mathrm{dx}\right]\\ = 16\mathrm{\pi }–2×\mathrm{}\left\{\sqrt{6}\mathrm{ }{\frac{2{\mathrm{x}}^{\frac{3}{2}}}{3}}_{0}^{2}+{\left[\frac{\mathrm{x}}{2}\sqrt{16-{\mathrm{x}}^{2}+8{\mathrm{sin}}^{-1}\frac{\mathrm{x}}{4}}\right]}_{2}^{4}\right\}\\ = 16\mathrm{\pi }–2×\mathrm{}\left\{{\frac{4\sqrt{12}}{3}}_{0}^{2}+{\left[4\mathrm{\pi }-\left(\sqrt{12}+\frac{8\mathrm{\pi }}{6}\right)\right]}_{2}^{4}\right\}\\ = 16\mathrm{\pi }–2×\left\{\frac{8\sqrt{3}}{3}-2\sqrt{3}+4\mathrm{\pi }+\frac{4\mathrm{\pi }}{3}\right\}\\ = 16\mathrm{\pi }–2×\left\{\frac{2\sqrt{3}}{3}+\frac{8\mathrm{\pi }}{3}\right\}\\ = 16\mathrm{\pi }\mathrm{–}\frac{4\sqrt{3}}{3}-\frac{16\mathrm{\pi }}{3}\\ =\mathrm{ }\frac{32\mathrm{\pi }}{3}\mathrm{–}\frac{4\sqrt{3}}{3}\\ =\mathrm{ }\frac{4}{3}\left(8\mathrm{\pi }-\sqrt{3}\right)\mathrm{sq}.\mathrm{units}\end{array}$

Q.10 Find the area of the region bounded by ellipse

$\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{4}\mathrm{ }=1$

.

Ans $\begin{array}{l}\frac{{\mathrm{x}}^{2}}{9}+\frac{{\mathrm{y}}^{2}}{4}\mathrm{ }=1 \mathrm{represents}\mathrm{ellipse}\mathrm{with}\mathrm{centre}\mathrm{at}\left(0,\mathrm{ }0\right) \mathrm{and}\mathrm{a}= 3,\mathrm{b}= 2.\\ \mathrm{The}\mathrm{required}\mathrm{area}= 4×\mathrm{}\frac{2}{3}\underset{0}{\overset{3}{\int }}\sqrt{3\frac{2}{}-{\mathrm{x}}^{2}}\mathrm{dx}\\ =\mathrm{ }4×\frac{2}{3}\mathrm{ }{{\left[\frac{\mathrm{x}}{2}\sqrt{{3}^{2}-{\mathrm{x}}^{2}+\frac{9}{2} {\mathrm{sin}}^{-1}\frac{\mathrm{x}}{3}}\right]}_{0}}^{3}\\ =\mathrm{ }4×\frac{2}{3}\left[\frac{9\mathrm{\pi }}{4}\right]\\ =\mathrm{ }6\mathrm{sq}.\mathrm{units}\end{array}$

Q.11 Using the method of integration find the area bounded by the curve |x| + |y| = 1.

Ans $\begin{array}{l}\left|\mathrm{x}\right|+\left|\mathrm{y}\right|\mathrm{ }=\mathrm{ }1\mathrm{ }\mathrm{or}\\ \mathrm{x}+\mathrm{y}= 1\mathrm{if}\mathrm{x},\mathrm{y}\ge 0\\ \mathrm{x}–\mathrm{y}= 1\mathrm{if}\mathrm{x}\ge 0,\mathrm{y}< 0\\ –\mathrm{x}+\mathrm{y}= 1 \mathrm{if}\mathrm{x}<0,\mathrm{y}\ge 0\\ –\mathrm{x}–\mathrm{y}= 1\mathrm{if}\mathrm{x},\mathrm{y}< 0\\ \mathrm{Area}\mathrm{of}\mathrm{shaded}\mathrm{region}= 4\mathrm{x}\left[\underset{0}{\overset{1}{\int }}\left(1-\mathrm{x}\right)\mathrm{dx}\right]\\ = 4\mathrm{x}\left[\underset{0}{\overset{1}{\int }}\left(1-\mathrm{x}\right)\mathrm{dx}\right]\\ =\mathrm{ }4\mathrm{x}{\frac{{\left(1-\mathrm{x}\right)}^{2}}{-2}}_{0}^{1}\\ =\mathrm{ }4\mathrm{x}\left[0+\frac{1}{2}\right]\mathrm{ }=\mathrm{ }\frac{4}{2}\\ =\mathrm{ }2\mathrm{sq}.\mathrm{units}.\end{array}$

Q.12 Find the area bounded by the curve x2 = 4y and the line x = 4y – 2.

Ans

$\begin{array}{l}\mathrm{x}= 4\mathrm{y}-2\mathrm{represents}\mathrm{a}\mathrm{straight}\mathrm{line}.\\ {\mathrm{x}}^{2}=4\mathrm{y} \mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\left(0,\mathrm{ }0\right).\\ \mathrm{From}\mathrm{x}= 4\mathrm{y}-2\mathrm{and}{\mathrm{x}}^{2}=4,\mathrm{we}\mathrm{get}\\ \mathrm{x}={\mathrm{x}}^{2}-2\\ \mathrm{or}{\mathrm{x}}^{2}-\mathrm{x}-2=0\\ \mathrm{or}\mathrm{ }\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)\mathrm{ }=\mathrm{ }0\\ \mathrm{or} \mathrm{x}=-1,\mathrm{ }2\\ \mathrm{The}\mathrm{required}\mathrm{area}=\underset{-1}{\overset{2}{\int }}\frac{\mathrm{x}+2}{4}\mathrm{dx}-\underset{-1}{\overset{2}{\int }}\frac{{\mathrm{x}}^{2}}{4}\mathrm{dx}\\ =\mathrm{ }\frac{1}{4}\mathrm{ }{\left[\frac{{\left(\mathrm{x}+2\right)}^{2}}{2}-\frac{{\mathrm{x}}^{3}}{3}\right]}_{1}^{2}\\ =\mathrm{ }\frac{1}{4}\left[\frac{15}{2}-\frac{9}{3}\right]=\frac{1}{4}\mathrm{ }\left[\frac{45-18}{6}\right]\mathrm{ }=\mathrm{ }\frac{1}{4}\left[\frac{27}{6}\right]\\ =\mathrm{ }\frac{9}{8}\mathrm{sq}.\mathrm{units}\end{array}$ Q.13 Using integration, find the area of the region bounded by the line 3y = 2x +4, x-axis and the line x = 1 and x = 3.

Ans $\begin{array}{l}3\mathrm{y}=2\mathrm{x}+4\mathrm{represents}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{intersecting}\mathrm{x}–\mathrm{axis}\mathrm{and}\mathrm{y}–\mathrm{axis}.\\ \mathrm{Also},\mathrm{x}= 1\mathrm{and}\mathrm{x}= 3\mathrm{are}\mathrm{straight}\mathrm{lines}\mathrm{parallel}\mathrm{to}\mathrm{y}–\mathrm{axis}.\\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{area}=\underset{1}{\overset{3}{\int }}\frac{2\mathrm{x}+4}{3}\\ =\mathrm{ }\frac{2}{3}\underset{1}{\overset{3}{\int }}\left(\mathrm{x}+2\right)\mathrm{dx}\\ =\mathrm{ }\frac{2}{3}\underset{1}{\overset{3}{\int }}\frac{{\mathrm{x}}^{2}}{2}+2{\mathrm{x}}_{1}^{3}\mathrm{ }\\ =\mathrm{ }\frac{2}{3}\left[\frac{9}{2}+6-\frac{1}{2}-2\right]\\ =\mathrm{ }\frac{16}{3}\mathrm{sq}.\mathrm{units}\end{array}$

Q.14

$\mathrm{Find}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{region}\mathrm{given}\mathrm{by}\left\{\mathrm{x},\mathrm{y}:{\mathrm{x}}^{2}\le \mathrm{y}\le \mathrm{x}\right\}\mathrm{using}\mathrm{integration}.$

Ans $\begin{array}{l}{\mathrm{x}}^{2}\mathrm{ }=\mathrm{ }\mathrm{y}\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\left(0,\mathrm{ }0\right).\\ \mathrm{y}=\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{origin}\mathrm{and}\mathrm{making}\mathrm{an}\mathrm{angle}\mathrm{of}45° \mathrm{with}\mathrm{the}\mathrm{x}–\mathrm{axis}.\\ \mathrm{Solving}{\mathrm{x}}^{2}=\mathrm{y}\mathrm{and}\mathrm{x}=\mathrm{y},\mathrm{we}\mathrm{get}\mathrm{x}= 1.\\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{area}=\left[\underset{0}{\overset{1}{\int }}\mathrm{xdx}-\underset{0}{\overset{1}{\int }}{\mathrm{x}}^{2}\mathrm{dx}\right]\\ = {\left[\frac{{\mathrm{x}}^{2}}{2}-\frac{{\mathrm{x}}^{3}}{3}\right]}_{0}^{1}\\ =\mathrm{ }\left[\frac{1}{2}-\frac{1}{3}\right]\\ =\mathrm{ }\frac{1}{6}\mathrm{sq}.\mathrm{units}\end{array}$

Q.15 Find the area of the region bounded by exponential function from x = 0 and x = 1.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{1}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{1}{\int }}{\mathrm{e}}^{\mathrm{x}}\mathrm{}\mathrm{dx}\\ ={{\left[{\mathrm{e}}^{\mathrm{x}}\right]}_{0}}^{1}\\ =\left[{\mathrm{e}}^{1}-{\mathrm{e}}^{0}\right]\\ =\left(\mathrm{e}-1\right) \mathrm{square}\mathrm{unit}\end{array}$

Q.16 Find the area of the region bounded by Line
x – y=0 , X-axis and ordinates x=1.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{1}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{1}{\int }}\mathrm{x}\mathrm{dx}\\ =\frac{1}{2}{{\left[{\mathrm{x}}^{2}\right]}_{0}}^{1}\\ =\frac{1}{2}\left[{\left(1\right)}^{2}-0\right]\\ =\frac{1}{2}\mathrm{square}\mathrm{unit}\end{array}$

Q.17

$\text{Find the area of the regionbounded by sine function from x = 0 and x =}\frac{\pi }{2}.$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{sinx}\mathrm{dx}\\ =-{{\left[\mathrm{cosx}\right]}_{0}}^{\frac{\mathrm{\pi }}{2}}\\ =-\left[\mathrm{cos}\frac{\mathrm{\pi }}{2}-\mathrm{coso}\right]\\ =-\left(0-1\right)=1\mathrm{}\mathrm{square}\mathrm{unit}\end{array}$

Q.18

$\text{Find the area of the region bounded by cosine function from x}=0\text{and x}=\frac{\pi }{2}.$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{cosx}\mathrm{dx}\\ ={{\left[\mathrm{sinx}\right]}_{0}}^{\frac{\mathrm{\pi }}{2}}\\ =\left[\mathrm{sin}\frac{\mathrm{\pi }}{2}-\mathrm{sino}\right]\\ =\left(1-0\right)=1\mathrm{}\mathrm{square}\mathrm{unit}\end{array}$

Q.19

$\text{Find the area of the region bounded by cosine function from x}=0\text{and x}=\pi \frac{\pi }{2}.$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\mathrm{cosx}\mathrm{dx}\\ ={{\left[\mathrm{sinx}\right]}_{0}}^{\frac{\mathrm{\pi }}{2}}\\ =\left[\mathrm{sin}\frac{\mathrm{\pi }}{2}-\mathrm{sino}\right]\\ =\left(1-0\right)=1\mathrm{}\mathrm{square}\mathrm{unit}\end{array}$

Q.20 Find the area of the region bounded by curve xy = 1 from x = 1 and x = e.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{1}{\overset{\mathrm{e}}{\int }}\mathrm{ydx}\\ =\underset{1}{\overset{\mathrm{e}}{\int }}\mathrm{}\frac{\mathrm{dx}}{\mathrm{x}}\left[\because \mathrm{xy}=1⇒\mathrm{y}=\frac{1}{\mathrm{x}}\right]\\ ={{\left[\mathrm{logx}\right]}_{1}}^{\mathrm{e}}\\ =\left[\mathrm{loge}-\mathrm{log}1\right]\\ =\left(1-0\right)=1\mathrm{}\mathrm{square}\mathrm{unit}\end{array}$

Q.21 Find the area of the region bounded by curve
y = x2, X-axis and ordinates x = 1.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{1}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{1}{\int }}{\mathrm{x}}^{2}\mathrm{}\mathrm{dx}\\ =\frac{1}{2}{{\left[{\mathrm{x}}^{3}\right]}_{0}}^{1}\\ =\frac{1}{2}\left[{\left(1\right)}^{3}-0\right]\\ =\frac{1}{2}\mathrm{square}\mathrm{unit}\end{array}$

Q.22 Find the area of the region bounded by line
y = 2x+1, X-axis, Y-axis and ordinates x = 1.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{1}{\int }}\mathrm{ydx}\\ =\underset{0}{\overset{1}{\int }}\left(2\mathrm{x}+1\right)\mathrm{}\mathrm{dx}\\ ={{\left[2\frac{{\mathrm{x}}^{2}}{2}+\mathrm{x}\right]}_{0}}^{1}\\ =\left[\left(1+1\right)-0\right]\\ =2\mathrm{}\mathrm{square}\mathrm{units}\end{array}$

Q.23 Find the area of the region bounded by curve y=f(x), ordinates x=a and x=b, as shown below.

Ans $\begin{array}{c}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{ydx}\\ =\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{}\mathrm{dx}\end{array}$

Q.24 Find the area of the region bounded by curve y = f(x), y = g(x) and ordinates x = a and x = b, as shown below.

Ans $\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\left[\mathrm{upper}-\mathrm{lower}\mathrm{}\mathrm{curve}\right]\mathrm{dx}\\ =\underset{\mathrm{a}}{\overset{\mathrm{b}}{\int }}\left[\mathrm{f}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{x}\right)\right]\mathrm{dx}\\ \end{array}$

Q.25 Find the area of the region bounded by parabola y2 = x and ordinates x=1.

Ans

$\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\mathrm{2}\underset{0}{\overset{1}{\int }}\mathrm{ydx}\\ =\mathrm{2}\underset{0}{\overset{1}{\int }}\sqrt{\mathrm{x}}\mathrm{}\mathrm{dx}\left[{\mathrm{y}}^{2}=\mathrm{x}⇒\mathrm{y}=\sqrt{\mathrm{x}}\right]\\ =\frac{4}{3}{{\left[{\left(\mathrm{x}\right)}^{\frac{3}{2}}\right]}_{0}}^{1}\\ =\frac{4}{3}\left[{\left(1\right)}^{\frac{3}{2}}-{\left(0\right)}^{\frac{3}{2}}\right]\\ =\frac{4}{3}\left[1-0\right]=\frac{4}{3}\mathrm{square}\mathrm{units}\end{array}$

Q.26 Find the area of the region bounded by parabola y2 = x and ordinates x = 1 and x = 4.

Ans

$\mathrm{We}\mathrm{have}\mathrm{parabola}{\mathrm{y}}^{2}=\mathrm{x}\mathrm{with}\mathrm{vertex}\mathrm{at}\mathrm{origin}.$ $\begin{array}{l}\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\mathrm{2}\underset{1}{\overset{4}{\int }}\mathrm{ydx}\\ =\mathrm{2}\underset{1}{\overset{4}{\int }}\sqrt{\mathrm{x}}\mathrm{}\mathrm{dx}\left[{\mathrm{y}}^{2}=\mathrm{x}⇒\mathrm{y}=\sqrt{\mathrm{x}}\right]\\ =\frac{4}{3}{{\left[{\left(\mathrm{x}\right)}^{\frac{3}{2}}\right]}_{1}}^{4}\\ =\frac{4}{3}\left[{\left(4\right)}^{\frac{3}{2}}-{\left(1\right)}^{\frac{3}{2}}\right]\\ =\frac{4}{3}\left[8-1\right]=\frac{28}{3}\mathrm{square}\mathrm{units}\end{array}$

Q.27 Find the area of the region bounded by curve y2 = x and straight line y = x.

Ans

$\begin{array}{l}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{curve}{\mathrm{y}}^{2}=\mathrm{x}\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\\ \mathrm{origin},\mathrm{symmetric}\mathrm{about}\mathrm{X}–\mathrm{axis}\mathrm{and}\mathrm{open}\mathrm{towards}\mathrm{right}\mathrm{hand}\mathrm{side}.\\ \mathrm{The}\mathrm{y}=\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{origin}\mathrm{inclined}\\ \mathrm{at}45° \left[\because \mathrm{m}=\mathrm{tan\theta }=1\right].\end{array}$ $\begin{array}{l}\mathrm{On}\mathrm{solving}\mathrm{these}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}\\ \mathrm{x}={\mathrm{x}}^{2}\\ ⇒\mathrm{x}\left(\mathrm{x}-1\right)=0\\ ⇒\mathrm{x}=0\mathrm{or}\mathrm{x}=1\\ \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{1}{\int }}\left[{\mathrm{Y}}_{\mathrm{upper}}-{\mathrm{Y}}_{\mathrm{lower}\mathrm{curve}}\right]\mathrm{dx}\\ =\underset{0}{\overset{1}{\int }}\left[\sqrt{\mathrm{x}}-\mathrm{x}\right]\mathrm{dx}\\ ={{\left[\frac{2}{3}{\left(\mathrm{x}\right)}^{\frac{3}{2}}-\frac{{\mathrm{x}}^{2}}{2}\right]}_{0}}^{1}\\ =\left[\left(\frac{2}{3}-\frac{1}{2}\right)-0\right]\\ =\frac{1}{6}\mathrm{}\mathrm{square}\mathrm{unit}\end{array}$

Q.28 Find the area of the region bounded by curve y=x2 and x=y2.

Ans

$\begin{array}{l}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{curve}{\mathrm{y}}^{2}=\mathrm{x}\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\\ \mathrm{origin},\mathrm{symmetric}\mathrm{about}\mathrm{X}–\mathrm{axis}\mathrm{and}\mathrm{open}\mathrm{towards}\mathrm{right}\mathrm{hand}\mathrm{side}.\\ \mathrm{The}\mathrm{curve}{\mathrm{x}}^{2}=\mathrm{y}\mathrm{represents}\mathrm{equation}\mathrm{of}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\\ \mathrm{at}\mathrm{origin},\mathrm{symmetric}\mathrm{about}\mathrm{Y}–\mathrm{axis}\mathrm{and}\mathrm{open}\mathrm{towards}\mathrm{upward}.\end{array}$ $\begin{array}{l}\mathrm{On}\mathrm{solving}\mathrm{these}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}\\ \mathrm{x}={\mathrm{x}}^{4}\\ ⇒\mathrm{x}\left({\mathrm{x}}^{3}-1\right)=0\\ ⇒\mathrm{x}=0\mathrm{or}\mathrm{x}=1\\ \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{1}{\int }}\left[{\mathrm{Y}}_{\mathrm{upper}}-{\mathrm{Y}}_{\mathrm{lower}\mathrm{curve}}\right]\mathrm{dx}\\ =\underset{0}{\overset{1}{\int }}\left[\sqrt{\mathrm{x}}-{\mathrm{x}}^{2}\right]\mathrm{dx}\\ ={{\left[\frac{2}{3}{\left(\mathrm{x}\right)}^{\frac{3}{2}}-\frac{{\mathrm{x}}^{3}}{3}\right]}_{0}}^{1}\\ =\left[\left(\frac{2}{3}-\frac{1}{3}\right)-0\right]\\ =\frac{1}{3}\mathrm{square}\mathrm{units}\end{array}$

Q.29

$\text{Find the area of the region bounded by curve}\phantom{\rule{0ex}{0ex}}\text{x =}\sqrt{y}\text{and straight line y = x+2;}\forall \text{x}\ge \text{0.}$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{know}\mathrm{that}\mathrm{curve}\mathrm{x}=\sqrt{\mathrm{y}}\left({\mathrm{x}}^{2}=\mathrm{y}\right)\mathrm{represents}\mathrm{parabola}\mathrm{with}\\ \mathrm{vertex}\mathrm{at}\mathrm{origin},\mathrm{symmetric}\mathrm{about}\mathrm{Y}–\mathrm{axis}\mathrm{and}\mathrm{open}\mathrm{towards}\mathrm{upwards}.\\ \because \mathrm{x}\ge 0\mathrm{for}\forall \mathrm{y}\in \mathrm{R}\\ \therefore \mathrm{There}\mathrm{is}\mathrm{no}\mathrm{portion}\mathrm{of}\mathrm{curve}\mathrm{that}\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{LHS}\mathrm{of}\mathrm{Y}–\mathrm{axis}.\\ \mathrm{We}\mathrm{have},\mathrm{y}=\mathrm{x}+2\\ \mathrm{at}\mathrm{x}=0,\mathrm{y}=2\\ \mathrm{at}\mathrm{y}=0,\mathrm{x}=-2\\ \mathrm{The}\mathrm{y}=\mathrm{x}+2\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{passing}\mathrm{through}\left(0,2\right)\mathrm{and}\left(-2,0\right)\\ \mathrm{inclined}\mathrm{at}45°. \left[\because \mathrm{m}=\mathrm{tan}45°=1\right]\end{array}$ $\begin{array}{l}\mathrm{On}\mathrm{solving}\mathrm{these}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}\\ {\mathrm{x}}^{2}=\mathrm{x}+2\\ ⇒{\mathrm{x}}^{2}-\mathrm{x}-2=0\\ ⇒\left(\mathrm{x}+1\right)\left(\mathrm{x}-2\right)=0\\ ⇒\mathrm{x}=-1\mathrm{or}\mathrm{x}=2\\ \because \mathrm{Shaded}\mathrm{portion}\mathrm{lies}\mathrm{in}\mathrm{x}\ge 0\mathrm{and}\mathrm{x}\le 2.\\ \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{bounded}\mathrm{region}=\underset{0}{\overset{2}{\int }}\left[{\mathrm{Y}}_{\mathrm{upper}}-{\mathrm{Y}}_{\mathrm{lower}\mathrm{curve}}\right]\mathrm{dx}\\ =\underset{0}{\overset{1}{\int }}\left[\left(\mathrm{x}+2\right)-{\mathrm{x}}^{2}\right]\mathrm{dx}\\ ={{\left[\frac{{\mathrm{x}}^{2}}{2}+2\mathrm{x}-\frac{{\mathrm{x}}^{3}}{3}\right]}_{0}}^{1}\\ =\left[\left(\frac{1}{2}+2-\frac{1}{3}\right)-0\right]\\ =2+\frac{1}{6}\mathrm{=}\frac{13}{6}\mathrm{sq}.\mathrm{units}\end{array}$

Q.30 Using the method of integration, find the area of the region bounded by inequation 2x+y 8, x+2y 8, x 0 and y 0.

Ans

$\begin{array}{l}\text{x+2y}\le \text{8}...\left(1\right)\\ \text{Let x+2y=8}\\ \text{Put x=0, we get y=4, that is point}\left(0,4\right).\\ \text{Put y=0, we get x=8, that is point}\left(8,0\right).\\ \text{We shall draw a line between these two points.}\\ \mathrm{Now},\text{put origin}\left(\mathrm{x}=0,\mathrm{y}=0\right)\text{in}\left(1\right)\text{, we get}\\ \text{0+0}\le \text{8}\\ ⇒0\le 8\\ \mathrm{which}\text{is true, so we will shade region containing origin.}\end{array}$ $\begin{array}{l}2\mathrm{x}+\mathrm{y}\le \mathrm{8}...\left(2\right)\\ \mathrm{Let}2\mathrm{x}+\mathrm{y}=8\\ \mathrm{Put}\mathrm{x}=0,\mathrm{we}\mathrm{get}\mathrm{y}=8,\mathrm{that}\mathrm{is}\mathrm{point}\left(0,8\right).\\ \mathrm{Put}\mathrm{y}=0,\mathrm{we}\mathrm{get}\mathrm{x}=4,\mathrm{that}\mathrm{is}\mathrm{point}\left(4,0\right).\\ \mathrm{We}\mathrm{shall}\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{through}\mathrm{these}\mathrm{two}\mathrm{points}.\\ \mathrm{Now},\mathrm{put}\mathrm{origin}\left(\mathrm{x}=0,\mathrm{y}=0\right)\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get}\\ 0+0\le \mathrm{8}\\ ⇒0\le 8\\ \mathrm{which}\mathrm{is}\mathrm{true},\mathrm{so}\mathrm{we}\mathrm{will}\mathrm{shade}\mathrm{region}\mathrm{containing}\mathrm{origin}.\\ \mathrm{Since}\mathrm{x}\ge 0\mathrm{and}\mathrm{y}\ge 0,\mathrm{represent}\mathrm{the}\mathrm{region}\mathrm{of}\mathrm{First}\mathrm{Quadrant}.\end{array}$ $\begin{array}{l}\mathrm{On}\mathrm{solving}\mathrm{the}\mathrm{equation},\\ 2\mathrm{x}+\mathrm{y}=8\mathrm{and}\mathrm{x}+2\mathrm{y}=8,\mathrm{we}\mathrm{get}\mathrm{x}=\frac{8}{3}\\ \mathrm{Thus},\mathrm{we}\mathrm{can}\mathrm{divide}\mathrm{the}\mathrm{shaded}\mathrm{region}\mathrm{in}\mathrm{two}\mathrm{parts}\mathrm{at}\mathrm{C}.\\ \mathrm{The}\mathrm{}\mathrm{bounded}\mathrm{}\mathrm{area}=\mathrm{Area}\mathrm{}\mathrm{under}\mathrm{}\mathrm{line}\mathrm{}\mathrm{AB}+\mathrm{Area}\mathrm{}\mathrm{under}\mathrm{}\mathrm{line}\mathrm{}\mathrm{BC}\\ =\underset{0}{\overset{\frac{8}{3}}{\int }}{\mathrm{Y}}_{\mathrm{AB}}\mathrm{dx}+\underset{\frac{8}{3}}{\overset{4}{\int }}{\mathrm{Y}}_{\mathrm{BC}}\mathrm{dx}\\ =\underset{0}{\overset{\frac{8}{3}}{\int }}\left(\frac{8-\mathrm{x}}{2}\right)\mathrm{dx}+\underset{\frac{8}{3}}{\overset{4}{\int }}\left(8-2\mathrm{x}\right)\mathrm{dx}\\ ={{\left[4\mathrm{x}-\frac{1}{2}.\frac{{\mathrm{x}}^{2}}{2}\right]}^{\frac{8}{3}}}_{0}+{{\left[8\mathrm{x}-2\frac{{\mathrm{x}}^{2}}{2}\right]}^{4}}_{\frac{8}{3}}\\ =\left[\left(\frac{32}{3}-\frac{16}{9}\right)-0\right]+\left[\left(32-16\right)-\left(\frac{64}{3}-\frac{64}{9}\right)\right]\\ =\frac{80}{9}+16-\frac{128}{9}\\ =16-\frac{48}{9}\\ =16-\frac{16}{3}\\ =\frac{32}{3}\mathrm{}\mathrm{sq}.\mathrm{}\mathrm{units}\end{array}$

Q.31 Find the area of the region bounded by the curve y2 = 4x and the line y = 4x – 2.

Ans

The given curve is: y2 = 4x …(i)

And the given line is: y = 4x – 2 …(ii)

On solving both the equations:

(4x – 2)2 = 4x

16x2 – 16x + 4 = 4x

16x2 – 20x + 4 = 0

Or 4x2 – 5x + 1 = 0

4x2 – 4x – x + 1 = 0

(4x – 1)(x – 1) = 0

Or x = 1/4, 1

And y = 4(1/4) – 2 or 4(1) – 2

Or y = –1 or 2

Therefore, the points of intersection are (1/4, –1) and (1, 2).

Coordinate of C:

y = 0 or 4x – 2 = 0 or x = ½

So, coordinates of C are (1/2, 0).

And coordinates of E are (1/4, 0).

The area bounded by the parabola y2 = 4x and line y = 4x – 2 is given below:   ## 1. Why should I refer to Chapter 8 Mathematics notes by Extramarks?

The Class 12 Mathematics Chapter 8 notes are prepared by professionals and subject experts at Extramarks who have years of experience in the subject. While solving the problems, if a student faces any difficulty , they can refer to the stepwise solutions and pictorial representation in the notes to understand each and every step. This will help them to strengthen their basics.

## 2. Which chapters are included in the Extramarks NCERT Solutions of Class 12 Mathematics?

Extramarks, an online learning platform, provides chapter-wise solutions such as the Class 12 Mathematics Chapter 8 notes to help students in their studies. All the crucial topics included in the syllabus are well-explained in a detailed and stepwise manner. Students of Class 12 are advised to study with the help of these academic notes to get a clear idea of the  exam pattern, marking system, weightage as well as important concepts from each chapter.

Chapter 1: Relations and Functions

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Applications of Derivatives

Chapter 7: Integrals

Chapter 8: Application of Integrals

Chapter 9: Differential Equations

Chapter 10: Vector Algebra

Chapter 11: Three Dimensional Geometry

Chapter 12: Linear programming

Chapter 13: Probability