The area of the region { ( x , y ) : | x − 1 | ≤ y ≤ 5 − x 2 } is equal to:

5 2 s i n − 1 ( 3 5 ) − 1 2

The area of the region { ( x , y ) : | x − 1 | ≤ y ≤ 5 − x 2 } is equal to:

5 2 s i n − 1 ( 3 5 ) − 1 2

NCERT Class 12 Maths Chapter 8 Solutions PDF – Application of Integrals

Q: 34. Using the method of integration, find the area of the region bounded by the lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

The given equation of the lines are 2x+y=4−−−−−−(1)3x−2y=6−−−−−(2)x−3y+5=0−−−−(3) ∴ Area of ?ABC=area(PCBQ)−area(?APC)−area(?AQB) =∫14y(3)dx−∫12y(1)dx−∫24y(2)dx−=∫14(x+5)3dx−∫12(4−2x)dx−∫24(3x−62dx) ∴ The point of intersection of the circle and the parabola is . A(12,)&C(12,−) Taking in first quadrant Area of (OABO)=area(OADO)+area(BADB)

Updated on Aug 5, 2025 15:56 IST

By Pallavi Pathak, Assistant Manager Content

Application of Integrals Class 12 Maths chapter deals with the study of a specific application of integrals to find the area between lines, the area under simple curves, the area between arcs of circles, ellipses, and parabolas. Areas under the curve are known as the elementary area.
Applications of Integrals Class 12 is created by the subject matter experts at Shiksha, hence it is reliable and accurate study material for students preparing for the CBSE Board exam and competitive exams like JEE Mains. The solution is created in a step-by-step format, which is easy to understand and gives students deep concept clarity.
Those who are looking for topic-wise revision PDFs with solutions must explore - Class 12 Maths Notes PDF for CBSE Exams.

Dive into Class 12 Integral Applications

Here are quick highlights of the Integral Applications Class 12 Maths:

For the curve y = f (x), x-axis and the lines x = a and x = b (b > a), area is - $Area = \int_{a}^{b} y d x = \int_{a}^{b} f (x) d x .$
For the curve x = φ (y), y-axis and the lines y = c, y = d, area is - $Area = \int_{c}^{d} x d y = \int_{c}^{d} ϕ (y) d y$

Class 12 Math Chapter 8 Application of Integrals : Key Topics, Weightage

If you are aiming to score high in an entrance exam such as JEE Mains, you should focus on practicing Class 12 Application of Integrals. It is a small chapter. Below are the topics covered in this chapter:

Exercise	Topics Covered
8.1	Introduction
8.2	Area under Simple Curves

Application of Integrals Class 12 Weightage in JEE Mains

Exam	Number of Questions	Weightage
JEE Main	1-2 questions	6-7%

Important Formulas for Class 12 Application of Integrals

Important Formulae for Class 12 Application of Integrals

Area under a curve:
$A = \int_{a}^{b} f (x) d x$
(For a function $y = f(x)$ between $x = a$ and $x = b$ )
Area between two curves:
$A = \int_{a}^{b} [f (x) - g (x)] d x$
(Where $f(x)$ and $g(x)$ are two functions and $f(x) \geq g(x)$ in the given interval)
For curves in terms of y:
$A = \int_{c}^{d} [g (y) - f (y)] d y$
(Where curves are given in the form $x = f(y)$ and $x = g(y)$ , and $g(y) \geq f(y)$ in the given interval)
Area Enclosed by the x-axis and the Curve:
$A = \int_{a}^{b} |f(x)| \,dx$

Class 12th Math Exercise 8.1 Solutions

Q1. Find the area of the region bounded by the curve y² = x the lines x=1, x = 4 and x-axis.

Q2. Find the area of the region bounded by the curve y² = 9x and the lines x=2,x=4 and x-axis in the first quadrant.

Q3. Find the area of the region bounded by the curve x² = 4y ,y=2,y=4 and y-axis is the first quadrant.

Q4. Find the area of the region bounded by the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ .

Commonly asked questions

6. Find the area of the region in the first quadrant enclosed by x-axis,line $x = \sqrt3y$ and the circle $x^{2} + y^{2} = 4$ .

Kindly go through the solution

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

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

So, equation of line AB is $y - 0 = \frac{5 - 0}{4 - 2} (x - 2)$

$= y = \frac{5}{2} (x - 2)$

Similarly equation of BC is

$\begin{array}{l} y - 5 = \frac{3 - 5}{6 - 4} (x - 4) \\ = y = 5 + (- x + 4) \\ y = 9 - x \end{array}$

And equation of AC is $y - 0 = \frac{3 - 0}{6 - 2} (x - 2)$

= $y = \frac{3}{4} (x - 2)$

$∴$ Area of $? A B C$

= $a r e a (? A B E) + a r e a (B C D E) - a r e a (? A C D)$

$\begin{array}{l} = \int_{2}^{4} y_{A B} d x + \int_{4}^{6} y_{B C} d x - \int_{2}^{6} y_{B C} d x \\ = \int_{2}^{4} \frac{5}{2} (x - 2) d x + \int_{4}^{6} (9 - x) d x - \int_{2}^{6} \frac{3}{4} (x - 2) d x \\ = \frac{5}{2} {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} + {[9 x - \frac{x^{2}}{2}]}_{4}^{6} - \frac{3}{4} {[\frac{x^{2}}{2} - 2 x]}_{2}^{6} \end{array}$

$\begin{array}{l} = \frac{5}{2} [(\frac{4^{2}}{2} - 2.4) - (\frac{2^{2}}{2} - 2.2)] + [(9.6 - \frac{6^{2}}{2}) - (9.4 - \frac{4^{2}}{2})] - \frac{3}{4} [(\frac{6^{2}}{2} - 2.6) - (\frac{2^{2}}{2} - 2.2)] \\ = \frac{5}{2} [(8 - 8) - (2 - 4)] + [54 - 18 - 36 + 8] - \frac{3}{4} [18 - 12 - 2 + 4] \\ = 5 + 8 - 6 = 7 u n i t^{2} \end{array}$

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

The given equation of the lines are

$\begin{array}{l} 2 x + y = 4 - - - - - - (1) \\ 3 x - 2 y = 6 - - - - - (2) \\ x - 3 y + 5 = 0 - - - - (3) \end{array}$

$∴$ Area of $? A B C = a r e a (P C B Q) - a r e a (? A P C) - a r e a (? A Q B)$

$\begin{array}{l} = \int_{1}^{4} y_{(3)} d x - \int_{1}^{2} y_{(1)} d x - \int_{2}^{4} y_{(2)} d x - \\ = \int_{1}^{4} \frac{(x + 5)}{3} d x - \int_{1}^{2} (4 - 2 x) d x - \int_{2}^{4} (\frac{3 x - 6}{2} d x) \end{array}$

$∴$ The point of intersection of the circle and the parabola is . $A (\frac{1}{2},) & C (\frac{1}{2}, -)$

Taking in first quadrant

Area of $(O A B O) = a r e a (O A D O) + a r e a (B A D B)$

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

Kindly go through the solution

5. Find the area of the region bounded by the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$

Kindly go through the solution

12. Choose the correct answer:

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

(A) π

(B) π/2

(C) π/3

(D) π/4

Given equation of the circle is

∴ option (A) is correct

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

Kindly go through the solution

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

Kindly go through the solution

4. Find the area of the region bounded by the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ .

Kindly go through the solution

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

The equation given circle is

$\begin{array}{l} 4 x^{2} + 4 y^{2} = 9 \\ \Rightarrow x^{2} + y^{2} = \frac{9}{4} \\ \Rightarrow x^{2} + y^{2} = {(\frac{3}{2})}^{2} \end{array}$

i.e, centre (0,0), radius $r = \frac{3}{2}$

since $x^{2} = 4 y$ intersect the circle

we can put $x^{2} = 4 y$ in $x^{2} + y^{2} = {(\frac{3}{2})}^{2}$

$\begin{array}{l} (4 y) + y^{2} = \frac{9}{4} \\ \Rightarrow y^{2} + 4 y - \frac{9}{4} = 0 \\ \Rightarrow 4 y^{2} + 16 y - 9 = 0 \\ \Rightarrow 4 y^{2} + 18 y - 2 y - 9 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 y (2 y + 9) - 1 (2 y + 9) = 0 \\ \Rightarrow (2 y + 9) (2 y - 1) = 0 \\ \Rightarrow y = \frac{- 9}{2} & y = \frac{1}{2} \end{array}$

$y = \frac{- 9}{2}, x^{2} = 4 \times (\frac{- 9}{2}) = - 18$ which is not possible or $x^{2}$ cannot be (-)ve

7. Find the area of the smaller part of the circle x² + y² = a² cut off by the line x = a/√2

Kindly go through the solution

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

Kindly go through the solution

\begin{array}{l} \Rightarrow {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{a} = {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{a}^{} \\ \Rightarrow \frac{2}{3} [a^{\frac{3}{2}} - 0] = \frac{2}{3} [4^{\frac{3}{2}} - a^{\frac{3}{2}}] \\ \Rightarrow a^{\frac{3}{2}} = 4^{\frac{3}{2}} - a^{\frac{3}{2}} \\ \Rightarrow 2 a^{\frac{3}{2}} = 4^{\frac{3}{2}} \end{array}

$\Rightarrow 4 a^{3} = 4^{3}$ (Squaring both sides)

$\Rightarrow a^{3} = 4^{2}$

$\Rightarrow a = 4^{\frac{2}{3}}$ (Taking cube on both sides)

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

Given that equation of

curve $y = x^{2}$

line $y = | x | = {x, i f x \geq 0 & - x, i f x ∠ 0}$

Since the line passes through A&B in Ist and IInd quadrants

the equation must satisfy

$\Rightarrow y = x^{2}$

$\Rightarrow x = x^{2}$ for Ist quadrant and

$\Rightarrow - x = x^{2}$ for IInd t quadrant

So, $\Rightarrow x^{2} - x = 0$ and $x^{2} + x = 0$

$\Rightarrow x (x - 1) = 0$ and $x (x + 1) = 0$

$\Rightarrow x = 0, 1$ $x = 0, - 1$

$x = 0, y = 0$

$x = 1, y = 1$ i.e, A has coordinate (1,1)

$x = - 1, y = 1$ i.e, B has coordinate (1,1)

Now, area of AODA = area (AOM)-area (ADOM)

$= \int_{0}^{1} y_{l i n e} d x - \int_{0}^{1} y_{c u r v e} d x$

$= \int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x = {[\frac{x^{2}}{2}]}_{0}^{1} - {[\frac{a^{3}}{3}]}_{0}^{1}$

$= \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} u n i t s^{2}$

$∴$ The required area of the region bounded by curve $y = x^{2}$ and line $y = | x |$ is $\frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} u n i t s^{2}$

10. Find the area bounded by the curve x = 4y and the line x = 4y -2

Given curve is $x^{2} = 4 y$ and the equation of line is $x = 4 y - 2$

The point of intersection of the curve and the line can be determine as follows.

Put, $x = 4 y - 2 \Rightarrow x + 2 = 4 y \Rightarrow y \frac{x + 2}{4}$

In $x^{2} = 4 y$ to determine value of x

i.e, $x^{2} = 4 \times \frac{(x + 2)}{4} = x + 2$

$\begin{array}{l} \Rightarrow x^{2} - x - 2 = 0 \Rightarrow x^{2} + x - 2 x - 2 = 0 \\ \Rightarrow x (x + 1) - 2 (x + 1) = 0 \\ \Rightarrow (x + 1) (x - 2) = 0 \end{array}$

$\Rightarrow x = 2$ and $x = - 1$

$x = 2$ , we have ${(2)}^{2} = 4 y \Rightarrow 4 - 4 y \Rightarrow y = 1$

And at $x = - 1$ we have ${(- 1)}^{2} = 4 y \Rightarrow y = \frac{1}{4}$

So, the coordinates A and B are (2,1) and ( $- 1, \frac{1}{4}$ )

$∴$ The required area before the line & the curve is area $B D^{ι} A B$ = area of trapezium (BNMAB)- area under curve BDA

$\begin{array}{l} = \int_{- 1}^{2} y_{l i n e} d x - \int_{- 1}^{2} y_{c u r v e} d x \\ = \int_{- 1}^{2} \frac{x + 2}{4} d a - \int_{- 1}^{2} \frac{x^{2}}{4} \\ = \frac{1}{4} \int_{- 1}^{2} x d a + \frac{2}{4} \int_{- 1}^{2} d x - \frac{1}{4} \int_{- 1}^{2} x^{2} d x \end{array}$

$\begin{array}{l} = \frac{1}{4} {[\frac{x^{2}}{2}]}_{- 1}^{2} + \frac{2}{4} {[x]}_{- 1}^{2} - \frac{1}{4} {[\frac{x^{3}}{3}]}_{- 1}^{2} \\ = \frac{1}{8} [2^{2} - {(- 1)}^{2}] + \frac{1}{2} [2 - (- 1)] - \frac{1}{12} [2^{3} - {(- 1)}^{3}] \end{array}$

$\begin{array}{l} = \frac{3}{8} + \frac{3}{2} - \frac{9}{12} = \frac{3}{8} + \frac{3}{2} - \frac{3}{4} = \frac{3 + 4 \times 3 - 2 \times 3}{8} \\ = \frac{3 + 12 - 6}{8} = \frac{9}{8} u n i t^{2} \end{array}$

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

The given equation of the curve is

$\begin{array}{l} y^{2} = 4 x \\ \Rightarrow y = \pm\sqrt3 \end{array}$

$\begin{array}{l} → y =+ \sqrt2x \end{array}$ in Ist quadrant

So, area of curve enclosed by $y^{2} = 4 x$

And $x = 3 = 2 \times area$ area (AOCA)

13. Choose the correct answer:

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

(A) 2

(B) 9/4

(C) 9/3

(D) 9/2

As $y = 3$ intersect $y^{2} = 4 x$ at Athen,

$\begin{array}{l} 3^{2} = 4 x \\ \Rightarrow x = \frac{9}{4} \end{array}$

$∴$ A has coordinate $(\frac{a}{4}, 3)$

Hence, area of curve = $\int_{y = 0}^{y = 3} x d y$

$\begin{array}{l} = \int_{0}^{3} \frac{y^{2}}{4} d y = {[\frac{y 3}{4 \times 3}]}_{0}^{3} \\ = \frac{3^{3}}{12} - \frac{0}{12} \\ = \frac{27}{12} = \frac{9}{4} u n i t^{2} \end{array}$

$∴$ Option (B) is correct

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

The equation of the given circle is

$= x^{2} + y^{2} = 1$ - (1)

$= (x-1) 2 + y^{2} = 1$ - (1) - (2)

Equation (1) is a circle with centre 0 (0,0) and radius 1. Equation (2) is a circle with centre c (1,0) and radius 1.

Solving (1) and (2)

16. Find the area of the region bounded by the curves y = x²+2, y = x,

x = 0 and x = 3

The equation of the curve is $y = x^{2} + 2 \Rightarrow x^{2} = y - 2$ - (1) and

lines are

$y = x$ - (2)

$x = 0$ - (3)

$x = 3$ - (4)

Equation (1)is a parabola with vertex (0,2)

Equation (2)is a straight line passing origin with shape = $t a n θ = 1 = θ = 4 5^{?}$

$∴$ The required area enclosed OBCDO = area (ODCAO)-area (OBAO)

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

Let A (-1,0),B(1,3) and C (3,2) be the vertices of a triangle ABC

So, equation of line AB is $y - 0 = \frac{3 - 0}{1 - (- 1)} (x - (- 1))$

$= y = \frac{3}{2} (x + 1)$ -------------(1)

Equation of line BC is $y - 3 = \frac{2 - 3}{3 - 1} (x - 1)$

$= y = \frac{- 1}{2} (x - 1) + 3 = \frac{- 1}{2} x + \frac{1}{2} + 3 = \frac{- x}{2} + \frac{7}{2}$ ---------------(2)

Equation of line AC is $y - 0 = \frac{2 - 0}{3 - (- 1)} [x - (- 1)]$

$= y = \frac{2}{4} (x + 1) = \frac{1}{2} (x + 1)$ ------------------------------(3)

$∴$ Area of $?$ ABC= area ( $? A B E$ ) $+ a r e a (B C D E)$ $- a r e a (? A C D)$

$\begin{array}{l} = \int_{- 1}^{1} y_{e q (1)} d x + \int_{1}^{3} y_{e q (2)} d x - \int_{- 1}^{3} y_{e q 3} d x \\ = \int_{- 1}^{1} \frac{3}{2} (x + 1) d x \int_{1}^{3} (\frac{- x}{2} + \frac{7}{2}) d x - \int_{- 1}^{1} \frac{1}{2} (x + 1) \\ = \frac{3}{2} {[\frac{x^{2}}{2} + x]}_{- 1}^{1} + \frac{1}{2} {[- \frac{x^{2}}{2} + 7 x]}_{1}^{3} d x - \frac{1}{2} {[\frac{x^{2}}{2} + x]}_{- 1}^{3} \end{array}$

$\begin{array}{l} = \frac{3}{2} [(\frac{1^{2}}{2} + 1^{2}) - (\frac{{(- 1)}^{2}}{2} + (π))] + \frac{1}{2} [(\frac{3^{2}}{2} + 7 \times 3) - (\frac{- 1^{2}}{2} + 7 \times 1)] - \frac{1}{2} [(\frac{3^{2}}{2} + 3) - (\frac{{(- 1)}^{2}}{2} + (- 1))] \\ = \frac{3}{2} [\frac{1}{2} + 1 - \frac{1}{2} + 1] + \frac{1}{2} [\frac{- 9}{2} + 21 + \frac{1}{2} - 7] - \frac{1}{2} [\frac{9}{2} + 3 - \frac{1}{2} + 1] \\ = \frac{3}{2} [2] + \frac{1}{2} [10] - \frac{1}{2} [8] = 3 + 5 - 4 = 8 - 4 = 4 u n i t^{2} \end{array}$

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

The given equation of the sides of triangle is

$y = 2 x + 1$ --------------------(1)

$y = 3 x + 1$ -------------------(2)

$x = 4$ -------------------------(3)

Solving eqn (1) and (2) for x & y we get

$\begin{array}{l} 3 x + 1 = 2 x + 1 \\ = 3 x - 2 x = 1 - 1 \\ = x = 0 \\ & y = 2 \times 0 + 1 = 1 \end{array}$

$∴$ The point of inersection of line (1)and (2)is A (0,1)

Putting x=4 in eq (1) and (2)we get,

$\begin{array}{l} y = 2 \times 4 + 1 = 8 + 1 = 9 \\ & y = 3 \times 4 + 1 = 12 + 1 = 13 \end{array}$

$∴$ The point of intersection of line (1)and (3) is B(4,9) and C (4,13)

Hence the required area enclosed ABC

$\begin{array}{l} = \int_{0}^{4} y_{l i n e (2)} d x - \int_{0}^{4} y_{l i n e (1)} d x \\ = \int_{0}^{4} [3 x + 1] d x - \int_{0}^{4} [2 x + 1] d x \\ = {[\frac{3 x^{2}}{2} + x]}_{0}^{4} - {[\frac{2 x^{2}}{2} + x]}_{0}^{4} \\ = [(\frac{3}{2} {(4)}^{2} + 4) - (\frac{3 \times 0^{2}}{2} + 0)] - [(4^{2} + 4) - (0^{2} +)] \\ = 24 + 4 - 20 = 8 u n i t^{2} \end{array}$

19. Choose the correct answer:

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

(A) 2 (π – 2)

(B) π – 2

(C) 2π – 1

(D) 2 (π + 2)

The equation of circle is $x^{2} + y^{2} = 4$ which has centre at (0,0) & radius,

$π = 2$

And the line $x + y = 2 = y = 2 - x$

The smaller area of circle is given by

Area (ABCA) area (BOAB) – area (BOA)

20. Choose the correct answer:

Area lying between the curves y² = 4x and y = 2x is

(A) 2/3

(B) 1/3

(C) 1/4

(D) 3/4

The given equation of the curve is $y^{2} = 4 x$ - (1) and

the line is $y = 2 x$ - (2)

Solving (1) and (2) for x and y

\begin{array}{l} {(2 x)}^{2} = 4 x \\ = 4 x^{2} = 4 x \\ = x^{2} - x = 0 \\ = x (x - 1) = 0 \end{array}

So, $x = 0 & x = 1$

for $x = 0$ we get $y = 2 \times 0 = 0$

for $x = 1$ , we get $y = 2 \times 1 = 2$

so, the point of intersection are (0,0)and (1,2)

area (DCAO)=area (DCABO)-area ( $? O A B$ )

21. 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

(i) Given of curve is $y = x^{2} \Rightarrow x^{2} = y$ and the equation are $x = 1 & x = 2 .$

$∴$ Area enclosed

$\begin{array}{l} = \int_{x = 1}^{x = 2} y d x \\ = \int_{1}^{2} x^{2} d x \\ = {[\frac{x^{3}}{3}]}_{1}^{2} = \frac{2^{3}}{3} - \frac{1^{3}}{3} \\ = \frac{8 - 1}{3} = \frac{7}{3} u n i t^{2} \end{array}$

(ii) Given equation of curve is $y = x^{4}$ and the lines are $x = 1 & x = 5$

So, area enclosed

$\begin{array}{l} = \int_{1}^{5} y d x \\ = \int_{1}^{5} x^{4} d x \\ = {[\frac{x^{5}}{5}]}_{1}^{5} = (\frac{5^{5} - 1^{5}}{5}) \\ = \frac{(3125 - 1)}{5} \\ = \frac{3124}{5} = 624.8 u n i t^{2} \end{array}$

22. Find the area between the curves y = x and y = x²

The given equation of the curve is $y = x^{2}$ --------(1)

and that of the line is $y = x$ ---------(2)

Solving eq (1) and (2)for x and y

$\begin{array}{l} x = x^{2} \\ = x^{2} - x = 0 \\ = x (x - 1) = 0 \\ = x = 0 & x = 1 \end{array}$

Where, $x = 0, y = 0^{2} = 0$

And when $x = 1, y = 1^{2} = 1$

$∴$ The point of intersection of the parabola $y = x^{2}$ and the line $y = x$

Is O(0,0) and B (1,1)

Hence, area between the curve and the line is

$a r e a (D C A O) = a r e a (? O A B) - a r e a (O A B O)$

$\begin{array}{l} = \int_{0}^{1} y_{l i n e} d x - \int_{0}^{1} y_{c u r v e} d x \\ = \int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x \\ = {[\frac{x^{2}}{2}]}_{0}^{1} - {[\frac{x^{3}}{3}]}_{0}^{1} \end{array}$

$\begin{array}{l} = \frac{1}{2} - \frac{1}{3} \\ = \frac{3 - 2}{6} = \frac{1}{6} u n i t^{2} \end{array}$

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

Given curve is $y = 4 x^{2} \Rightarrow x^{2} = \frac{1}{4} y$ - (1)

$x = 0$ i.e, y-axis and y=4 and y=1

Hence, the required area in Ist quadrant i.e, area ABCD = $\int_{y = 1}^{y = 4} x d y$

24. Sketch the graph of y = |x + 3| and evaluate

Given equation of lines is $y = | x + 3 |$ -------(1)

The point (x,y)satisfying (1)are

Hence plotting the above in graph we get

Now, $\int_{- 6}^{0} | x + 3 | d x = \int_{- 6}^{- 3} | x + 3 | d x + \int_{- 3}^{0} | x + 3 | d x$

We know that, $\begin{array}{l} \int_{- 6}^{0} | x + 3 | d x = \int_{- 6}^{- 3} | x + 3 | d x + \int_{- 3}^{0} | x + 3 | d x \\ y = | x + 3 | = {\begin{cases} x + 3, i f, x + 3 \geq 0 \Rightarrow x \geq - 3 \\ - (x + 3), i f, x + 3 ∠ 0 \Rightarrow x ∠ - 3 \end{cases}} \end{array}$

So, $\int_{- 6}^{0} | x + 3 | d x = \int_{- 6}^{- 3} - (x + 3) d x + \int_{- 3}^{0} (x + 3) d x$

$\begin{array}{l} = - {[\frac{x^{2}}{2} + 3 x]}_{- 6}^{- 3} + {[\frac{x^{2}}{2} + 3 x]}_{- 3}^{0} \\ = - {[\frac{{(- 3)}^{2}}{2} + 3 (- 3)] - [\frac{{(- 6)}^{2}}{2} + 3 (- 6)]} + {[\frac{0^{2}}{2} + 3 \times 0] - [\frac{{(- 3)}^{2}}{2} + 3 \times (- 3)]} \\ = - {\frac{9}{2} - 9 - \frac{36}{2} + 18} + {- \frac{9}{2} + 9} \\ = - \frac{9}{2} + 9 + 18 - 18 - \frac{9}{2} + 9 \\ = 9 u n i t^{2} \end{array}$

25. Find the area bounded by the curve y = sin x between x = 0 and x = 2π

The given equation of the curve is $y = s i n x$

$∴$ The required area bounded by the curve

$\begin{array}{l} = \int_{0}^{π} y d x + \int_{π}^{2 π} y d x \\ = \int_{0}^{π} s i n x d x + \int_{π}^{2 π} s i n x d x \\ = | {[- c o s x]}_{0}^{π} | + | {[- c o s x]}_{0}^{π} | \\ = | - [c o s x - c o s θ] | + | - [c o s 2 π - c o s π] | \\ = | - [- 1 - 1] | + | - [1 + 1] | \\ = | 2 | + | - 2 | = 2 + 2 = 4 u n i t^{2} \end{array}$

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

The equation of the parabola is $y^{2} = 4 a^{2}$ -----------(1)

and that of line is $y = m x$ ------(2)

The Point of intersection of(1)and (2) is given by

$\begin{array}{l} {(m x)}^{2} = 4 a x \\ \Rightarrow m^{2} x^{2} - 4 a x = 0 \\ \Rightarrow x (m^{2} x - 4 a) = 0 \\ \Rightarrow x = 0 & x = \frac{4 a}{m^{2}} \end{array}$

For, $x = 0, y^{2} = 4 a \times 0 \to y = 0$ i.e, O(0,0)

For, $x = \frac{4 a}{m^{2}}, y^{2} = 4 a \times \frac{4 a}{m^{2}} \to y = \frac{4 a}{m}$ (in first quadrant)

i.e, $A (\frac{4 a}{m^{2}}, \frac{4 a}{m})$

Hence, the required area enclosed by the curve and the lines is

$a r e a (D A C O) = a r e a (O C A B O) - a r e a (? O A B)$

27. Find the area enclosed by the parabola 4y = 3x² and the line 2y = 3x + 12

The given equation of parabola is $4 y = 3 x^{2} \Rightarrow x^{2} = \frac{4}{3} y$ ------------(1)

And the line is $2 y = 3 x + 12 \Rightarrow y = \frac{3}{2} x + 6$ ----------------------(2)

Solving (1) and (2) for x and y,

$\begin{array}{l} x^{2} = \frac{4}{3} (\frac{3}{2} x + 6) = 2 x + 8 \\ \Rightarrow x^{2} - 2 x - 8 = 0 \\ \Rightarrow x^{2} - 4 x + 2 x - 8 = 0 \\ \Rightarrow x (x - 4) + 2 (x - 4) = 0 \\ \Rightarrow (x - 4) (x + 2) = 0 \\ \Rightarrow x = 4 & x = - 2 \end{array}$

At, $x = 4, y = \frac{3}{2} \times 4 + 6 = 6 + 6 = 12$

And $x = - 2, y = \frac{3}{2} \times (- 1) + 6 = - 3 + 6 = 3$

Thus, the point of intersection of (1)&(2)are $A (4, 12) & B (- 2, 3)$

$∴$ Area of the enclosed region (BOAB)

=area (CBAD) – area (OADC)

$\begin{array}{l} = \int_{- 2}^{4} y_{l i n e} d x - \int_{- 2}^{4} y_{c u r v e} d x \\ = \int_{- 2}^{4} (\frac{3}{2} x + 6) d x - \int_{- 2}^{4} \frac{3 x^{2}}{4} d x \\ = {[\frac{3}{2} \frac{x^{2}}{2} + 6 x]}_{- 2}^{4} - {[\frac{3}{4} \times \frac{x^{3}}{3}]}_{- 2}^{4} \\ = [(\frac{3}{4} \times 4^{2} + 6 \times 4) - (\frac{3}{4} {(- 2)}^{2} + 6 \times (- 2))] - \frac{1}{4} [4^{3} - {(- 2)}^{3}] \\ = [12 + 24 - 3 + 12] - \frac{1}{4} [64 + 8] \\ = 45 - 18 = 27 u n i t^{2} \end{array}$

28. Find the area of the smaller region bounded by the ellipse and the line

Given equation of the ellipse is $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ Which as major axis aling x- axis and that of the line is $\frac{x}{3} + \frac{y}{2} = 1$ which has x and y intercepts at 3 and 2respectively.

Required area of enclosed region is area $a r e a (B C A B) = a r e a (O B A O) ? a r e a (A B O A)$

29. Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and the line $\frac{x}{a} + \frac{y}{b} = 1$

The Given equation of the ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

And the equation of the line in $\frac{x}{a} + \frac{y}{b} = 1$

With x and y intersept a and b

So, required area of the enclosed region is

$a r e a (B C A B) = a r e a (O B C A O) - a r e a (? O B A)$

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

The given equation of the parabola is $x^{2} = y$ ---------(1)

and that the line is $y = x + 2$ --------------(2)

Solving (1) and (2) for x and y

$\begin{array}{l} x^{2} = x + 2 \\ \to x^{2} - x - 2 = 0 \\ = x^{2} + x - 2 x - 2 = 0 \\ = x (x + 1) - 2 (x + 1) = 0 \\ = (x + 1) (x - 2) = 0 \\ = x = - 1 & x = 2 \end{array}$

When $x = - 1, y = {(- 1)}^{2} = 1$

And $x = 2, y = 2^{2} = 4$

$∴$ The point of intersection of the parabola and the lines $A (- 1, 1) & B (2, 4)$

Hence the required area enclosed region is $e a (A O B A) = a r e a (D A B C D) - a r e a (D A O B C D)$

$\begin{array}{l} = \int_{- 1}^{2} y_{l i n e} d x - \int_{- 1}^{2} y_{c u r v e} d x \\ = \int_{- 1}^{2} (x + 2) d x - \int_{- 1}^{2} x^{2} d x \\ = {[\frac{x^{2}}{2} 2 x]}_{- 1}^{2} - {[\frac{x^{3}}{3}]}_{- 1}^{2} \end{array}$

$\begin{array}{l} = [(\frac{2^{2}}{2} 2.2) - (\frac{{(- 1)}^{2}}{2} + 2 (- 1))] - [\frac{2^{3}}{3} - \frac{{(- 1)}^{3}}{3}] \\ = [2 + 4 - \frac{1}{2} + 2] - [\frac{8 + 1}{3}] = \frac{15}{2} - 3 = \frac{9}{2} u n i t^{2} \end{array}$

31. Using the method of integration, find the area enclosed by the curve |x| + |y| = 1

[Hint: the required region is bounded by lines x + y = 1, x – y = 1, – x + y = 1 and – x – y = 11]

Given equation of the curve is $| x | + | y | = 1$ , which can be break down into each quadrant .

For Ist quadrant, $| x | = x, | y | = y$

i.e., $x + y = 1$ - (1)

Similarly for IInd, IIIRd nad IVth quadrant

$- x + y = 1$ - (2)

$- x - y = 1$ - (3)

$x - y = 1$ - (4)

We draw the above focus lines on a graph and find the area enclosed which is a square.

$∴$ Required area $(? A B C D) = 4 \times a r e a (? A O B)$ .

$\begin{array}{l} = 4 \times \int_{0}^{1} y d x \\ = 4 \int_{0}^{1} (- x) d x \\ = 4 {[x - \frac{x^{2}}{2}]}_{0}^{1} \\ = 4 [1 - \frac{1}{2}] \\ = 4 \times \frac{1}{2} \\ = 2 u n i t^{2} \end{array}$

32. Find the area bounded by curves {(x, y) : y ≥ x2 and y = |x|}.

Given that equation of

curve $y = x^{2}$

line $y = | x | = {x, i f x \geq 0 & - x, i f x ∠ 0}$

Since the line passes through A&B in Ist and IInd quadrants

the equation must satisfy

$\Rightarrow y = x^{2}$

$\Rightarrow x = x^{2}$ for Ist quadrant and

$\Rightarrow - x = x^{2}$ for IInd t quadrant

So, $\Rightarrow x^{2} - x = 0$ and $x^{2} + x = 0$

$\Rightarrow x (x - 1) = 0$ and $x (x + 1) = 0$

$\Rightarrow x = 0, 1$ $x = 0, - 1$

$x = 0, y = 0$

$x = 1, y = 1$ i.e, A has coordinate (1,1)

$x = - 1, y = 1$ i.e, B has coordinate (1,1)

Now, area of AODA = area (AOM)-area (ADOM)

$= \int_{0}^{1} y_{l i n e} d x - \int_{0}^{1} y_{c u r v e} d x$

$= \int_{0}^{1} x d x - \int_{0}^{1} x^{2} d x = {[\frac{x^{2}}{2}]}_{0}^{1} - {[\frac{a^{3}}{3}]}_{0}^{1}$

$= \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} u n i t s^{2}$

$∴$ The required area of the region bounded by curve $y = x^{2}$ and line $y = | x |$ is $\frac{1}{6} + \frac{1}{6} = \frac{2}{6} =$ $\frac{1}{3} u n i t s^{2}$

35. Find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.

The given equation of curve $y^{2} \leq 4 x$ i.e, $y^{2} = 4 x$ - (1) is a parabola and

$4 x^{2} + 4 y^{2} \leq 9 \Rightarrow x^{2} + y^{2} \leq \frac{9}{4} \Rightarrow x^{2} + y^{2} = {(\frac{3}{2})}^{2}$ - (2) is a circle

With centre (0,0)and radius

Solving (1) and (2) for x and y,

$\begin{array}{l} x^{2} + 4 x = \frac{9}{4} \\ = x^{2} + 4 x - \frac{9}{4} = 0 \\ = 4 x^{2} + 16 x - 9 = 0 \end{array}$

36. Choose the correct answer:

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

Given is $y = x^{3}$ and the ines $x = - 2 & x = 1$

For $y = x^{3}$

$\begin{array}{l} a r e a (O A B) = \int_{0}^{1} y d x \\ = \int_{0}^{1} x^{3} d x \\ = {[\frac{x^{4}}{4}]}_{0}^{1} = \frac{1}{4} \\ a r e a (O D C) = \int_{- 2}^{0} y d x \\ = \int_{- 2}^{0} x^{3} d x \\ = | {[\frac{x^{4}}{4}]}_{- 2}^{0} | = | [\frac{0^{4}}{4} - \frac{{(- 2)}^{4}}{4}] | = 4 \end{array}$

$∴$ Total area of the bounded region $= \frac{1}{4} + 4$

$= \frac{17}{4} u n i t^{2}$

37. Choose the correct answer:

The area bounded by the curve y = x|x|, axis and the ordinates x = –1 and x = 1 is given by:

[Hint: y = x² if x > 0 and y = –x² if x < 0]

(A) 0

(B) 1/3

(C) 2/3

(D) 4/3

The given curve is $y = x | x |$

$= {(x . x i f x \geq 0) (x . (- x) i f x \leq 0)} = {(x^{2} i f, x \geq 0) (- x^{2} i f, x \leq 0)}$

Which is in the form of a parabola nad the lines are $x = - 1 & x - a x i s$

At $x = 1 > 0, y = 1^{2} = 1$

At $x = - 1 < 0, y = - 1^{2} = - 1$

Shaded area of the Ist quadrant

$\begin{array}{l} = \int_{0}^{1} y d x = \int_{0}^{1} x^{2} d x \\ = {[\frac{x^{3}}{3}]}_{0}^{1} \\ = \frac{1}{3} \end{array}$

Shaded area of the IInd quadrant

$\begin{array}{l} = \int_{- 1}^{0} y d x = \int_{- 1}^{0} x^{2} d x \\ = {[\frac{- x^{3}}{- 3}]}_{- 1}^{0} \\ = \frac{1}{3} \end{array}$

$∴$ Total area of the enclosed region $= \frac{1}{3} + \frac{1}{3}$

$= \frac{2}{3} u n i t^{2}$

$∴$ Option (c) is correct.

38. Choose the correct answer:

The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is

a. $\frac{4}{3} (4 π -)$

b. $\frac{4}{3} (4 π +)$

c. $\frac{4}{3} (8 π -)$

d. $\frac{4}{3} (4 π +)$

The given area of the circle is $x^{2} + y^{2} = 16 - - - - - - - (1)$ is a circle with centre (0,0) and radius, $π = 4$ and the parabola is $y^{2} = 6 x$ -------------(2)

Solving (1) and (2) for x and y.

$\begin{array}{l} x^{2} + 6 x = 16 \\ = x^{2} + 6 x - 16 = 0 \\ = x^{2} + 8 x - 2 x - 16 = 0 \\ = x (x + 8) - 2 (x + 8) = 0 \\ = (x + 8) (x - 2) = 0 \\ = x = - 8 & x = 2 \end{array}$