Here is a NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals. This solutions covers all questions of Maths Chapter 8 Application of Integrals Class 12 as per CBSE Board guidelines from the latest NCERT book for class 12 maths. Following topics and sub-topics in Class 12 Maths Chapter 8 Application of Integrals are covered.

8.1 Introduction

8.2 Area under Simple Curves

8.2.1 The area of the region bounded by a curve and a line

8.3 Area between Two Curves.

## Application of Integrals NCERT Solutions – Class 12 Maths

### NCERT Solutions for Class 12 Maths Chapter 8 – Free PDF Download

**Exercise 8.1 :** Solutions of Questions on Page Number** : 365**

**Q1 : ****Find the area of the region bounded by the curve y ^{2} = x and the lines x = 1, x = 4 and the x-axis.**

**Answer :**

The area of the region bounded by the curve, y

^{2}= x, the lines,x = 1 and x = 4, and the x-axis is the area ABCD.

**Q2 :**

**Find the area of the region bounded by y**

^{2}= 9x, x = 2, x = 4 and the x-axis in the first quadrant.**Answer :**

The area of the region bounded by the curve, y

^{2}= 9x, x = 2, and x = 4, and the x-axis is the area ABCD.

**Q3 :**

**Find the area of the region bounded by x**

^{2}= 4y, y = 2, y = 4 and the y-axis in the first quadrant.**Answer :**

The area of the region bounded by the curve, x

^{2}= 4y, y = 2, and y = 4, and the y-axis is the area ABCD.

**Q4 :**

**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

**Q5 :**

**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 =

**Q6 :**

**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

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

**Q7 :**

**Find the area of the smaller part of the circle x**

^{2}+ y^{2}= a^{2}cut off by the line**Answer :**

The area of the smaller part of the circle, x

^{2}+ y

^{2}= a

^{2}, 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, x

^{2}+ y

^{2}= a

^{2}, cut off by the line, is units.

**Q8 :**

**The area between x = y**

^{2}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 parabola and

*x*= 4 into two equal parts.

∴ Area OAD = Area ABCD

∴ 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

**Q9 :**

**Find the area of the region bounded by the parabola y = x**

^{2}and**Answer :**

The area bounded by the parabola, x

^{2}= 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, x

^{2}= 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

**Q10 :**

**Find the area bounded by the curve x**

^{2}= 4y and the line x = 4y – 2**Answer :**

The area bounded by the curve, x

^{2}= 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 =

**Q11 :**

**Find the area of the region bounded by the curve y**

^{2}= 4x and the line x = 3**Answer :**

The region bounded by the parabola, y

^{2}= 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.

**Q12 :**

**Area lying in the first quadrant and bounded by the circle x**

^{2}+ y^{2}= 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.

**Q13 :**

**Area of the region bounded by the curve y**

^{2}= 4x, y-axis and the line y = 3 is**A. 2**

**B.**

**C.**

**D.**

**Answer :**

The area bounded by the curve, y

^{2}= 4x, y-axis, and y = 3 is represented as

Thus, the correct answer is B.

**Exercise 8.2 :**Solutions of Questions on Page Number

**: 371**

**Q1 : Find the area of the circle 4x**

^{2}+ 4y^{2}= 9 which is interior to the parabola x^{2}= 4y**Answer :**

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, 4x

^{2}+ 4y

^{2}= 9, and parabola, x

^{2}= 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

**Q3 : Find the area of the region bounded by the curves y = x**

^{2}+ 2, y = x, x = 0 and x = 3**Answer :**

The area bounded by the curves, y = x

^{2}+ 2, y = x, x = 0, and x = 3, is represented by the shaded area OCBAO as

Then, Area OCBAO = Area ODBAO – Area ODCO

**Q4 :**

**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

**Q5 :**

**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)

**Q6 :**

**Smaller area enclosed by the circle x**

^{2}+ y^{2}= 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, x

^{2}+ y

^{2}= 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.

**Q7 :**

**Area lying between the curve y**

^{2}= 4x and y = 2x is**A.**

**B.**

**C.**

**D.**

**Answer :**

The area lying between the curve, y

^{2}= 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.

**Exercise Miscellaneous :**Solutions of Questions on Page Number

**: 375**

**Q1 :**

**Find the area under the given curves and given lines:**

**(i) y = x**

^{2}, x = 1, x = 2 and x-axis**(ii) y = x**

^{4}, 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

**Q2 :**

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

^{2}**Answer :**

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x

^{2}, is A (1, 1).

We draw AC perpendicular to x-axis.

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

**Q3 :**

**Find the area of the region lying in the first quadrant and bounded by y = 4x**

^{2}, x = 0, y = 1 and y = 4**Answer :**

The area in the first quadrant bounded by y = 4x

^{2}, x = 0, y = 1, and y = 4 is represented by the shaded area ABCDA as

**Q4 :**

**Sketch the graph of and evaluate**

**Answer :**

The given equation isThe 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,

**Q5 :**

**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

**Q6 :**

**Find the area enclosed between the parabola y**

^{2}= 4ax and the line y = mx**Answer :**

The area enclosed between the parabola, y

^{2}= 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)

**Q7 :**

**Find the area enclosed by the parabola 4y = 3x**

^{2}and the line 2y = 3x + 12**Answer :**

The area enclosed between the parabola, 4y = 3x

^{2}, 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)

**Q8 :**

**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)

**Q9 :**

**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)

**Q10 :**

**Find the area of the region enclosed by the parabola x**

^{2}= y, the line y = x + 2 and x-axis**Answer :**

The area of the region enclosed by the parabola, x

^{2}= y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as

The point of intersection of the parabola, x

^{2}= y, and the line, y = x + 2, is A (-1, 1) and C(2, 4).

**Q11 :**

**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

**Q12 :**

**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.

= 7 units

**Q14 :**

**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 â€“ 2y = 6 … (2)

And, x â€“ 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)

**Q15 :**

**Find the area of the region**

**Answer :**

The area bounded by the curves, , is represented as

The points of intersection of both the curves are.

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about x-axis.

∴ Area OABCO = 2 × Area OBC

Area OBCO = Area OMC + Area MBC

Therefore, the required area is units

**Q16 :**

**Area bounded by the curve y = x**

^{3}, the x-axis and the ordinates x = -2 and x = 1 is**A. – 9**

**B.**

**C.**

**D.**

**Answer :**

**Q17 :**

**The area bounded by the curve, x-axis and the ordinates x = -1 and x = 1 is given by**

**[Hint: y = x**

^{2}if x > 0 and y = -x2 if x < 0]**A. 0**

**B.**

**C.**

**D.**

**Answer :**

Thus, the correct answer is C.

**Q18 :**

**The area of the circle x**

^{2}+ y^{2}= 16 exterior to the parabola y^{2}= 6x is**A.**

**B.**

**C.**

**D.**

**Answer :**

The given equations are

x

^{2}+ y

^{2}= 16 … (1)

y

^{2}= 6x … (2)

Area bounded by the circle and parabola

Area of circle = π (r)

^{2}

= π (4)

^{2}

= 16π units

Thus, the correct answer is C.

**Q19 :**

**The area bounded by the y-axis, y = cos x and y = sin x when**

**A.**

**B.**

**C.**

**D.**

**Answer :**

The given equations are

y = cos x … (1)

And, y = sin x … (2)

Required area = Area (ABLA) + area (OBLO)

Integrating by parts, we obtain

Thus, the correct answer is B.