**Chapter 5 Application of Definite Integration Miscellaneous Exercise 5**

## Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

**I. Choose the correct option from the given alternatives:**

**Question 1.The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by**

(a) 12 sq units

(b) 8 sq units

(c) 25 sq units

(d) 32 sq units

Answer:

(a) 12 sq units

**Question 2.**

**Question 3.**

**Question 5.**

**Question 6.The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is**
(a) 1 sq unit

(b) 2 sq units

(c) 3 sq units

(d) 4 sq units

**Answer:**

(d) 4 sq units

**Question 7.**

**Question 9.**

**Question 11.**

**Question 13.**

**Question 15.**

**Question 16.**

**Question 18.**

**Question 20.**

**(II) Solve the following:**

**Question 1.Find the area of the region bounded by the following curve, the X-axis and the given lines:(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2(ii) y = sin x, x = 0, x = π(iii) y = sin x, x = 0, x = π3Solution:**

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.

Two bounded regions and are obtained. Both the regions have equal areas.

∴ required area =

**Question 2.Find the area of the circle = 9, using integration.Solution:**

By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.

Clearly, for this region, the limits of integration are 0 and 3.

∴ area of the circle = 4 (area of the region OABO)

**Question 3.Solution:**

By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.

Clearly, for this region, the limits of integration are 0 and 5.

From the equation of the ellipse

∴ area of the ellipse = 4(area of the region OABO)

**Question 4.Find the area of the region lying between the parabolas:Solution:**

(i)

For finding the points of intersection of the two parabolas, we equate the values of from their equations.

(ii)

(iii)

**Question 5.Find the area of the region in the first quadrant bounded by the circle = 4 and the X-axis and the line x = y√3.Solution:**

**Question 6.Find the area of the region bounded by the parabola = x and the line y = x in the first quadrant.Solution:**

To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.

∴ = y

∴ – y = 0

∴ y(y – 1) = 0

∴ y = 0 or y = 1

When y = 0, x = 0

When y = 1, x = 1

∴ the points of intersection are O(0, 0) and A(1, 1).

Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO

Now, area of the region OCADO = area under the parabola = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1

Area of the region OBADO = area under the line y = x between x = 0 and x = 1

**Question 7.Find the area enclosed between the circle + = 1 and the line x + y = 1, lying in the first quadrant.Solution:**

Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)

Now, area of the region OACBO = area under the circle + =1 between x = 0 and x = 1

Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1

**Question 8.Find the area of the region bounded by the curve (y – 1 = 4(x + 1) and the line y = (x – 1).Solution:**

The equation of the curve is (y – 1 = 4(x + 1)

This is a parabola with vertex at A (-1, 1).

To find the points of intersection of the line y = x – 1 and the parabola.

Put y = x – 1 in the equation of the parabola, we get

When x = 0, y = 0 – 1 = -1

When x = 8, y = 8 – 1 = 7

∴ the points of intersection are B (0, -1) and C (8, 7).

To find the points where the parabola (y – 1 = 4(x + 1) cuts the Y-axis.

Put x = 0 in the equation of the parabola, we get

(y – 1 = 4(0 + 1) = 4

∴ y – 1 = ±2

∴ y – 1 = 2 or y – 1 = -2

∴ y = 3 or y = -1

∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).

To find the point where the line y = x – 1 cuts the X-axis.

Put y = 0 in the equation of the line, we get

x – 1 = 0

∴ x = 1

∴ the line cuts the X-axis at the point G (1, 0).

Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO

Now, area of the region BFAB = area under the parabola (y – 1 = 4(x + 1), Y-axis from y = -1 to y = 3

Since, the area cannot be negative,

Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG

**Question 9.Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.Solution:**

The equation of the line is

**Question 10.Find the area of the region bounded by the curve y = 4, Y-axis and the lines y = 1, y = 4.Solution:**

By symmetry of the parabola, the required area is 2 times the area of the region ABCD.