NCERT Solutions Class 12 Mathematics Chapter 8 Application of Integrals

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. 
Answer : 

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

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

Question. Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1 
Answer : 
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 
Answer : 
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. 
Answer : 
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
Answer : 
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.
Answer : 
The line, x = a, divides the area bounded by the parabola and x = 4 into two equal parts. 
Area OAD = Area ABCD 

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|
Answer :  
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 = 4y – 2
Answer : 
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
Answer : 
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 x = 2 is 
(A) π 
(B) π/2
(C) π/3
(D) π/4
Answer : 
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 
Answer : 
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 
Answer : 
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
Answer : 
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 = x² + 2, y = x, x = 0 and x = 3
Answer : 
The area bounded by the curves, y = x² + 2, y = x, x = 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).
Answer : 
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 x = 4.
Answer : 
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)
Answer : 
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
Answer : 
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 = x², x = 1, x = 2 and x-axis
(ii) y = x⁴, x = 1, x = 5 and x –axis
Answer : 
(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.
Answer : 
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 = 4x², x = 0, y = 1 and y = 4.
Answer : 
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.
Answer : 
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 x between x = 0 and x = 2π.
Answer : 
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.
Answer : 
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
Answer : 
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.
Answer : 
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.
Answer : 
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.
Answer : 
The area of the region enclosed by the parabola, x² = y , 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].
Answer : 
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|}.
Answer : 
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).
Answer : 
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 – 3y + 5 = 0
Answer : 
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}
Answer : 
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
Answer : 

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 
Answer : 

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