Here we are going through the ncert solutions for class 12 maths chapter 8 – application of integrals so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions more easily

# Application Of Integrals

# EXERCISE 8.1

#### Question 1:

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

#### ANSWER:

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

#### Question 2:

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

#### ANSWER:

The area of the region bounded by the curve, y2 = 9x, x = 2, and x = 4, and the x-axis is the area ABCD.

Question 3:

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

#### ANSWER:

The area of the region bounded by the curve, x2 = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.

#### Question 4:

Find the area of the region bounded by the ellipse

#### 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 of OAB

Therefore, area bounded by the ellipse = 4 × 3π = 12π units

#### Question 5:

Find the area of the region bounded by the ellipse

#### 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 =

#### Question 6:

Find the area of the region in the first quadrant enclosed by x-axis, line and the circle

#### ANSWER:

The area of the region bounded by the circle, , and the x-axis is the area OAB.

The point of intersection of the line and the circle in the first quadrant is .

Area OAB = Area ΔOCA + Area ACB

Area of OAC

Area of ABC

Therefore, required area enclosed = 3√2 + π3 − 3√2 = π3 square units32 + π3 - 32 = π3 square units

#### Question 7:

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line

#### ANSWER:

The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, , is the area ABCDA.

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, , is units.

#### Question 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.

#### 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

From (1) and (2), we obtain

Therefore, the value of a is .

#### Question 9:

Find the area of the region bounded by the parabola y = x2 and

#### ANSWER:

The area bounded by the parabola, x2 = y,and the line,, 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).

Area of OACO = Area ΔOAM – Area OMACO

Area of ΔOAM

Area of OMACO

⇒ Area of OACO = Area of ΔOAM – Area of OMACO

Therefore, required area = units

#### Question 10:

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 .

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO … (1)

Then, Area OBCO = Area OMBC – Area OMBO

Similarly, Area OACO = Area OLAC – Area OLAO

Therefore, required area =

#### Question 11:

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.

The area OACO is symmetrical about x-axis.

∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

#### Question 12:

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.

C.

D.

#### ANSWER:

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

Thus, the correct answer is A.

#### Question 13:

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

A. 2

B.

C.

D.

#### ANSWER:

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

Thus, the correct answer is B.

#### EXERCISE 8.2

#### Question 1:

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.

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.

Therefore, Area OBCO = Area OMBCO – Area OMBO

=∫2√0(9−4x2)4−−−−−−√dx−∫2√0x24dx=∫02(9-4x2)4dx-∫02x24dx

=∫2√0(32)2−x2−−−−−−−−√dx−14∫2√0x2dx=∫02322-x2dx-14∫02x2dx

=(x2(32)2−x2−−−−−−−−√+98sin−12x3)2√0−14(x33)2√0=x2322-x2+98sin-12x302-14x3302

=2√4+98sin−122√3−112(2–√)3=24+98sin-1223-11223

=122√+98sin−122√3−132√=122+98sin-1223-132

=162√+98sin−122√3=162+98sin-1223

=12[2√6+94sin−122√3]=1226+94sin-1223

Therefore, the required area OBCDO is units

#### Question 2:

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 + y 2 = 1, is represented by the shaded area as

On solving the equations, (x – 1)2 + y2 = 1 and x2 + y 2 = 1, we obtain the point of intersection as Aand B.

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 .

Therefore, required area OBCAO = units

#### Question 3:

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 by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

#### Question 4:

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 (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 – 4) = 4 units

#### Question 5:

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).

It can be observed that,

Area (ΔACB) = Area (OLBAO) –Area (OLCAO)

Question 6:

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

It can be observed that,

Area ACBA = Area OACBO – Area (ΔOAB)

Thus, the correct answer is B.

#### Question 7:

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

A.

B.

C.

D.

#### ANSWER:

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

The points of intersection of these curves are O (0, 0) and A (1, 2).

We draw AC perpendicular to x-axis such that the coordinates of C are (1, 0).

∴ Area OBAO = Area (OCABO) – Area (ΔOCA)

square units

Thus, the correct answer is B.

#### Miscellaneous Exercise

#### Question 1:

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:

The required area is represented by the shaded area ADCBA as

The required area is represented by the shaded area ADCBA as

#### Question 2:

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

#### ANSWER:

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x2, is A (1, 1).

We draw AC perpendicular to x-axis.

∴ Area (OBAO) = Area (ΔOCA) – Area (OCABO) … (1)

#### Question 3:

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

Area of ABCDA = ∫41 x dy =∫41 y√2 dy [as, y = 4x2] =12∫41y√ dy =12×23[y3/2]41 =13[(4)3/2 − (1)3/2] =13(8 − 1) =13×7 =73 square unitsArea of ABCDA = ∫14 x dy =∫14 y2 dy as, y = 4x2 =12∫14y dy =12×23y3/214 =1343/2 - 13/2 =138 - 1 =13×7 =73 square units

#### Question 4:

Sketch the graph of and evaluate

#### ANSWER:

The given equation is

The corresponding values of x and y are given in the following table.

x | – 6 | – 5 | – 4 | – 3 | – 2 | – 1 | 0 |

y | 3 | 2 | 1 | 0 | 1 | 2 | 3 |

On plotting these points, we obtain the graph of as follows.

It is known that,

#### Question 5:

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

∴ Required area = Area OABO + Area BCDB

#### Question 6:

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 .

We draw AC perpendicular to x-axis.

∴ Area OABO = Area OCABO – Area (ΔOCA)

#### Question 7:

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 8:

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

#### ANSWER:

The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

#### Question 9:

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

#### ANSWER:

The area of the smaller region bounded by the ellipse, , and the line, , is represented by the shaded region BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

#### Question 10:

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 OACO as

The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).

Area of OACO = ∫2−1(x + 2) dx − ∫2−1 x2 dx⇒Area of OACO = [x22 + 2x]2−1 − 13[x3]2−1⇒Area of OACO = [{(2)22+2(2)} − {(−1)22+2(−1)}] − 13[(2)3 − (−1)3]⇒Area of OACO = [2 + 4 − (12−2)] − 13(8 + 1)⇒Area of OACO = 6 + 32 − 3⇒Area of OACO = 3 + 32 = 92 square units Area of OACO = ∫-12x + 2 dx - ∫-12 x2 dx⇒Area of OACO = x22 + 2x-12 - 13x3-12⇒Area of OACO = 222+22 - -122+2-1 - 1323 - -13⇒Area of OACO = 2 + 4 - 12-2 - 138 + 1⇒Area of OACO = 6 + 32 - 3⇒Area of OACO = 3 + 32 = 92 square units

Question 11:

Using the method of integration find the area bounded by the curve

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

#### ANSWER:

The area bounded by the curve, , is represented by the shaded region ADCB as

The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0).

It can be observed that the given curve is symmetrical about x-axis and y-axis.

∴ Area ADCB = 4 × Area OBAO

Question 12:

Find the area bounded by curves

#### ANSWER:

The area bounded by the curves, , is represented by the shaded region as

It can be observed that the required area is symmetrical about y-axis.