# NCERT Solutions for Class 12 Maths Application of Integrals

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Application of Integrals Exercise 8.2*.

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**CBSE Class 12**^{th}Maths## NCERT Solutions for Class 12 Maths Application of Integrals

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 1.**

Find the area of the circle 4x^{2} + 4y^{2} = 9 which is interior to the parabola x^{2} = 4y.

**NCERT Solutions:**

Required area is represented by the shaded area OBCDO.

On solving the given equation of circle 4x^{2} + 4y^{2} = 9 and the equation of parabola x^{2} = 4y, we get the point of intersection as

Putting x^{2} = 4y in 4x^{2} + 4y^{2} = 9, then

Putting y = -9/2 in x^{2} = 4y then.

Here this is NOT possible as square of any number is always positive.

Putting y = 1/2 in x^{2} = 4y then

Thus, the point of intersection as .

Since the required area is symmetrical about y-axis, then

Area OBCDO = 2(Area OBCO). Now, we draw BM perpendicular to OA. Thus, the coordinate of M is (√2, 0).

Area OBCO = Area OMBCO – Area OMBO

Therefore the required area is

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 2.**

Find the area bounded by curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1.

**NCERT Solutions:**

Required area bounded by the given curves (x – 1)^{2} + y^{2} = 1 and x^{2} + y^{2} = 1 is represented by the shaded area OBCAO as below.

Now, On solving the given equations

(x – 1)^{2} + y^{2} = 1 …(1)

and x^{2} + y^{2} = 1 …(2),

we get

The point of intersection as

From the diagram, the required area is symmetrical about x-axis.

Thus, Area OBCAO = 2 x (Area OCAO). We join AB that intersect OC at M and hence AM is perpendicular to OC.

Thus, the coordinates of M is M(1/2, 0).

Therefore, the required area OBCAO is units.

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 3.**

Find the area of the region bounded by the curves y = x^{2} + 2, y = x, x = 0 and x = 3.

**NCERT Solutions:**

Required area bounded by the given curves y = x^{2} + 2, y = x, x = 0 and x = 3 is represented by the shaded area OCBAO as below.

Thus,

Area OCBAO = Area ODBAO – Area ODCO.

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 4.**

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

**NCERT Solutions:**

Required area is given by the shaded triangle ABC whose coordinates are A(-1, 0), B(1, 3) and C(1, 3).

Since the equation of the line AB is

Since the equation of the line BC is

Since the equation of the line AC is

Now,

Area (Triangle ABC)

= Area (Triangle ALB) + Area (Quadrilateral BLMC) – Area (Triangle AMC)

= (3 + 5 – 4)

= 4 units.

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 5.**

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

**NCERT Solutions:**

Required area is given by the shaded triangle ABC whose the equations of the sides are y = 2x + 1, y = 3x + 1 and x = 4 as below.

On solving the given equations y = 2x + 1, y = 3x + 1 and x = 4, we get the vertices of the triangle ABC as A(0, 1), B(4, 13) and C(4, 9).

From the figure, we get that

Area (Triangle ABC) = Area(OLBAO) – Area (OLCAO)

Choose the correct answer in the following exercises 6 and 7.

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 6.**

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)

**NCERT Solutions:**

(B)

Required area bounded by the circle x^{2} + y^{2} = 4 and the line x + y = 2 is the shaded area ACBA as below.

Now, from the figure, we get that

Area (ACBA) = Area (OACBO) – Area (OAB)

**NCERT Solutions for Class 12 Maths Application of Integrals ****Exercise**** 8.2: Ques No 7.**

Area lying between the curves y^{2} = 4x and y = 2x is

(A) 2/3

(B) 1/3

(C) 1/4

(D) 3/4

**NCERT Solutions:**

(B)

Required area bounded by the curves y^{2} = 4x and y = 2x is the shaded area OBAO as below.

Now, on solving the given equations, we get the point of intersection as O(0, 0) and A(1, 2).

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

Thus, Area (OBAO) = Area (OCA) – Area (OCABO)