# Class 12 Maths NCERT Solutions for Chapter 8 Application of Integrals Exercise 8.1

### Integrals Exercise 8.1 Solutions

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

**Solution**

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.

Area of ABCD = ∫

= (2/3) [8 - 1]

= 14/3 units

_{1}^{4}y dx**=**∫_{1}^{4}√x dx= (2/3) [8 - 1]

= 14/3 units

**2. Find the area of the region bounded by**

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

The area of the region bounded by the curve,

*y*

^{2}= 9

*x*,

*x*= 2, and

*x*= 4, and the x - axis is the area ABCD.

Area of ABCD = ∫

= ∫

= 2[8 - 2√2]

= (16 - 4√2) units

_{2}^{4}y dx= ∫

_{2}^{4}3√x dx= 2[8 - 2√2]

= (16 - 4√2) units

**3. Find the area of the region bounded by**

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

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.

Area of ABCD = ∫

= ∫

= 2∫

_{2}^{4}y dx= ∫

_{2}^{4}2√y dy= 2∫

_{2}^{4}√y dy= (4/3)[8 - 2√2]

= (32 - 8√2)/3 units

**4. Find the area of the region bounded by the ellipse x**

^{2}/16 + y^{2}/9 = 1**Solution**

The given equation of the ellipse, x

^{2}/16 + y^{2}/9 = 1 , can be represented asIt 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 = ∫

_{2}^{4}y dxTherefore, area bounded by the ellipse = 4 × 3Ï€ = 12Ï€ units

**5. Find the area of the region bounded by the ellipse x**

^{2}/4 + y^{2}/9 = 1**Solution**

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× (3Ï€/2) = 6Ï€ units

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

**6. Find the area of the region in the first quadrant enclosed by x-axis, line x = √3 y and the circle x**

^{2}+ y^{2}= 4.**Solution**

The area of the region bounded by the circle, x

^{2}+ y^{2}= 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 Î”OCA + 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 x

^{2}+ y^{2}= 4 in thefirst quadrant = √3/2 + Ï€/2 - √3/2 = Ï€/3 units

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

^{2}+ y^{2}= a^{2}cut off by the line x = a/√2**Solution**

The area of the smaller part of the circle, x

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

^{2}+ y^{2}= a^{2}, cut off by the line, x = a/√2, is [(a^{2}/2)(Ï€/2 - 1)] units.**8. 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*.**Solution**

The line, x = a, divides the area bounded by the 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

From (1) and (2), we obtain

⇒ 2.(a)

^{3/2}= 8

⇒ (a)

⇒ a = (4)

^{3/2}= 4⇒ a = (4)

^{2/3}Therefore, the value of a is (4)

^{2/3}.**9. Find the area of the region bounded by the parabola y = x**

^{2}and y = |x| .**Solution**

The area bounded by the parabola, x

^{2}= y , and the line, y = |x|, can be represented asThe 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).Therefore, required area = 2[1/6] = 1/3 sq units

**10. Find the area bounded by the curve**

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

The area bounded by the curve, x

^{2}= 4*y*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

Therefore, required area = (5/6 + 7/24) = 9/8 units

We draw AL and BM perpendicular to x - axis.

Similarly, Area OACO = Area OLAC - Area OLAO

Therefore, required area = (5/6 + 7/24) = 9/8 units

**11. Find the area of the region bounded by the curve**

*y*^{2}= 4*x*and the line*x*= 3**Solution**

The region bounded by the parabola,

*y*^{2}= 4*x,*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 8√3 units.

**12. Area lying in the first quadrant and bounded by the circle**

(A) Ï€

(B) Ï€/2

(C) Ï€/3

(D) Ï€/4

*x*^{2}+*y*^{2}= 4 and the lines*x*= 0 and*x*= 2 is(A) Ï€

(B) Ï€/2

(C) Ï€/3

(D) Ï€/4

**Solution**

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.

**13. Area of the region bounded by the curve**

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

(B) 9/4

(C) 9/3

(D) 9/2

(B) 9/4

(C) 9/3

(D) 9/2

**Solution**

The area bounded by the curve,

*y*^{2}= 4x, y - axis, and y = 3 is represented asThus, the correct answer is B.