# NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals

## Chapter 8 Application of Integrals Class 12 Mathematics NCERT Solutions

Class 12 Mathematics students should refer to the following NCERT questions with answers for Chapter 8 Application of Integrals in Class 12. These NCERT Solutions with answers for Class 12 Mathematics will come in exams and help you to score good marks

### Chapter 8 Application of Integrals NCERT Solutions Class 12 Mathematics

Exercise 8.1

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

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

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

Question. Find the area of the region bounded by the ellipse x2/16 + y2/9 = 1
The given equation of the ellipse, x2 /16 + y2 /9 = 1 , can be represented as

It can be observed that the ellipse is symmetrical about x - axis and y - axis.
∴ Area bounded by ellipse = 4× Area of OAB
Area of OAB = ∫24 y dx

Question. Find the area of the region bounded by the ellipse  x2/4 + y2/9 = 1
The given equation of the ellipse can be represented as

It can be observed that the ellipse is symmetrical about x - axis and y - axis.
∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse = 4× (3π/2) = 6π units

Question. Find the area of the region in the first quadrant enclosed by x-axis, line x = √3 y  and the circle x2 + y2 = 4.
The area of the region bounded by the circle, x2 + y2 = 4, x = √3 y, and the x - axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is (√3, 1) .
Area OAB = Area ΔOCA + Area ACB
Area of OAC = (1/2) × OC × AC = (1/2) × √3 × 1 = √3/2  ...(1)

Therefore, area enclosed by x - axis, the line x = √3 y, and the circle x2 + y2 = 4 in the
first quadrant = √3/2  + π/2 - √3/2 =  π/3 units

Question. Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2
The area of the smaller part of the circle, x2 + y2 = a2 , cut off by the line, x = a/√2, is the area ABCD.

It can be observed that the area ABCD is symmetrical about x - axis.
∴ Area ABCD = 2 × Area ABC

Therefore, the area of smaller part of the circle , x2 + y2 = a2 , cut off by the line, x = a/√2, is [(a2 /2)(π/2 - 1)] units.

Question. The area between x = y2 and x = 4 is divided into two equal parts by the line x = a, find the value of a.
The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts.

It can be observed that the given area is symmetrical about x - axis.
∴ Area OED  = Area EFCD

Question. Find the area of the region bounded by the parabola y = x2 and y = |x|
The area bounded by the parabola, x2 = y , and the line, y = |x|, can be represented as

The given area is symmetrical about y - axis,
∴ Area OACO = Area ODBO
The point of intersection of parabola, x2 = y, and line, y = x, is A(1, 1).

Question. Find the area bounded by the curve x2 = 4y and the line x = 4– 2
The area bounded by the curve, x2 = 4y and line, x = 4y - 2, is represented by the shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.
Coordinates of point  A are (-1, 1/4).
Coordinates of point B are (2, 1)
We draw AL and BM perpendicular to x - axis.
Similarly, Area OACO = Area OLAC - Area OLAO

Question. Find the area of the region bounded by the curve y2 = 4x and the line x = 3
The region bounded by the parabola, y2 = 4x, and the line, x = 3, is the area OACO.

Question. Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and = 2 is
(A) π
(B) π/2
(C) π/3
(D) π/4

The area bounded by the circle and the lines, x = 0 and x = 2, in the first quadrant is represented as

Question. 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

The area bounded by the curve, y2 = 4x, y - axis, and y = 3  is represented as

Exercise 8.2

Question. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y , we obtain the point of intersection as B(√2, 1/2) and D(-√2, 1/2).
It can be observed that the required area is symmetrical about y - axis.
Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are (√2, 0).
Therefore, Area OBCO = Area OMBCO - Area OMBO

Question. Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y 2 = 1
The area bounded by the curves, (x - 1)2 + y2 = 1 and x2 + y2 = 1, is represented by the shaded area as

On solving the equations, (x - 1)2 + y2 = 1 and x2 + y2 = 1, we obtain the point of intersections as A (1/2, √3/2) and B(1/2, √3/2)
It can be observed that the required area is symmetrical about x - axis.
∴ Area OBCAO = 2 × Area OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are (1/2, 0)
⇒ Area OCAO = Area OMAO + Area MCAM

Question. Find the area of the region bounded by the curves y = x+ 2, xx = 0 and x = 3
The area bounded by the curves, y = x+ 2, xx = 0 and x = 3, is represented b the shaded area OCBAO as

Question. Using integration finds the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
BL and CM are drawn perpendicular to x - axis.
It can be observed in the following figure that,
Area (Δ ACB) = Area (ALBA) + Area (BLMCB) - Area (AMCA) ...(1)

Question. Using integration find the area of the triangular region whose sides have the equations y = 2x +1, y = 3x + 1 and = 4.
The equations of sides of the triangle are y = 2x + 1, y = 3x + 1, and x = 4
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C(4, 9)

Question. 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 smaller area enclosed by the circle, x2 + y2 = 4 and the line x + y = 2 , is represented by the shaded area ACBA as

Question. Area lying between the curve y2 = 4x and y = 2x is
A. 2/3
B. 1/3
C. 1/4
D. 3/4

The area lying between the curve, y2  = 4x and y = 2x , is represented by the shaded area OBAO as

Miscellaneous Solutions

Question. Find the area under the given curves and given lines:
(i) y = x2x = 1, x = 2 and x-axis
(ii) y = x4x = 1, x = 5 and x –axis

(i) The required area is represented by the shaded area ADCBA as

(ii) The required area is represented by the shaded area ADCBA as

Question. Find the area between the curves y = x and y = x2.
The required area is represented by the shaded area OBAO as

Question. Find the area of the region lying in the first quadrant and bounded by y = 4x2x = 0, y = 1 and = 4.
The area in the first quadrant bounded by y = 4x2 , x = 0, y = 1 and y = 4 is represented by the shaded area ABCDA as

Question. Sketch the graph of y = |x + 3| and evaluate ∫-60 |x + 3| dx.
The given equation is y = |x + 3|
The corresponding values of x and y are given in the following table :

Question. Find the area bounded by the curve y = sin between x = 0 and x = 2π.
The graph of y = sin x can be drawn as

= [ - cosπ + cos 0] + |- cos 2π + cos π|
= 1 + 1+ |(-1-1)|
= 2 + |-2|
= 2 + 2 = 4 units

Question. Find the area enclosed between the parabola y2 = 4ax and the line y mx.
The area enclosed between the parabola, y2 = 4ax and the line y mx, is represented by the shaded area OABO as

The points of intersection of both the curves are (0, 0) and (4a/m2, 4a/m)
We draw AC perpendicular to x - axis.
Area OABO = Area OCABO - Area (ΔOCA)

Question. Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
The area enclosed between the parabola, 4y = 3x2 , and the line, 2y = 3x + 12, is represented by the shaded area OBAO as

The points of intersection of the given curves are A(-2, 3) and (4, 12).
We draw AC and BD perpendicular to x - axis.
Area OBAO = Area CDBA - (Area ODBO + Area OACO)

Question. Find the area of the smaller region bounded by the ellipse x2 /9 + y2 /4 and the line x/3 + y/2 = 1.
The area of the smaller region bounded by the ellipse, x2/9 + y2/4 = 1, and the line x/3 + y/2 = 1 , is represented by the shaded region BCAB as

Question. Find the area of the smaller region bounded by the ellipse x2 /a2  + y2 /b2  = 1 and the line x/a + y/b = 1.
The area of the smaller region bounded by the ellipse, x2 /a2  + y2 /b2  = 1 and the line, x/a + y/b = 1, is represented by the the shaded region BCAB as

Question. Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and x-axis.
The area of the region enclosed by the parabola, x2 = , the line, y = x + 2 and x - axis is represented by the shaded region OABCO as

Question. Using the method of integration find the area bounded 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 = 1].
The area bounded by the curve, |x| + |y| = 1, is represented by the shaded region ADCB as

Question. Find the area bounded by curves {(x, y) : y ≥ x2  and y = |x|}.
The area bounded by the curves, {(x, y) : y ≥ x2 and y = |x|}, is represented by the shaded region as

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

Equation of line segment AB is
y - 0 = [(5 - 0)/(4 - 2)] (x - 2)
⇒ 2y = 5x - 10
⇒ y = (5/2) (x - 2)                   ...(1)
Equation of line segment BC is
y - 5 = [(3 - 5)(6 - 4)] (x - 4)
⇒ 2y - 10 = -2x + 8
⇒ 2y = -2x + 18
⇒ y = -x + 9                   ...(2)
Equation of line segment CA is
y - 3 = 0 - 3 / 2 - 6 (x-6)
⇒ -4y + 12 = -3x + 18
⇒ 4y = 3x - 6
⇒ y = (3/4)(x - 2)                    ...(3)
Area (ΔABC) = Area (ABLA) + Area (BLMCB) - Area(ACMA)

Question. Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3+ 5 = 0
The given equations of lines are
2x + y= 4                                  ...(1)
3x -2 y = 6                                ...(2)
And, x - 3y + 5 = 0                   ...(3)

The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x - axis.
Area (ΔABC) = Area (ALMCA) - Area (ALB) - Area(CMB)

Question. Find the area of the region {(x, y) : y2  ≤ 4x, 4x2  + 4y2  ≤ 9}
The area bounded by the curves, {(x, y) : y2  ≤ 4x, 4x2  + 4y2  ≤ 9}, is represented as

Question. 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

Question. Choose the correct answer The area bounded by the curve y = x | 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

Question. Choose the correct answer The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is
A. 4/3 (4π - √3)
B. 4/3 (4π + √3)
C. 4/3 (8π - √3)
D. 4/3 (4π + √3)

The given equations are
x2 + y2 = 16 ...(1)
y2 = 6x ...(2)

Question. The area bounded by the y-axis, y = cos x and y = sin x when 0 <= x <= π2
(A) 2 ( 2 −1)
(B) √2-1
(C) √2+1
(D) √2

The given equation are
y = cos x ...(1)
And, y = sin x ...(2)

 NCERT Solutions Class 12 Mathematics Chapter 1 Relations and Functions
 NCERT Solutions Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions
 NCERT Solutions Class 12 Mathematics Chapter 3 Matrices
 NCERT Solutions Class 12 Mathematics Chapter 4 Determinants
 NCERT Solutions Class 12 Mathematics Chapter 5 Continuity and Differentiability
 NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives
 NCERT Solutions Class 12 Mathematics Chapter 7 Integrals
 NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals
 NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations
 NCERT Solutions Class 12 Mathematics Chapter 10 Vector Algebra
 NCERT Solutions Class 12 Mathematics Chapter 11 Three Dimensional Geometry
 NCERT Solutions Class 12 Mathematics Chapter 12 Linear Programming
 NCERT Solutions Class 12 Mathematics Chapter 13 Probability

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### NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals

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