**Chapter 5 Application of Definite Integration Ex 5.1**

## Chapter 5 Application of Definite Integration Ex 5.1

**1. Find the area of the region bounded by the following curves, X-axis, and the given lines:**

**(i) y = 2x, x = 0, x = 5.Solution:**

**(v) xy = 2, x = 1, x = 4.Solution:**

**(vi) = x, x = 0, x = 4.Solution:**

The required area consists of two bounded regions A1 and A2 which are equal in areas.

For = x, y = √x

Required area =

**(vii) = 16x, x = 0, x = 4.Solution:**

The required area consists of two bounded regions A1 and A2 which are equal in areas.

For = x, y = √x

Required area =

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

**(i) = 16x and its latus rectum.Solution:**

Comparing = 16x with = 4ax, we get

4a = 16

∴ a = 4

∴ focus is S(a, 0) = (4, 0)

For = 16x, y = 4√x

Required area = area of the region OBSAO

= 2 [area of the region OSAO]

**(ii) y = 4 – and the X-axis.Solution:**

The equation of the parabola is y = 4 –

∴ = 4 – y

i.e. (x – 0 = -(y – 4)

It has vertex at P(0, 4)

For points of intersection of the parabola with X-axis,

we put y = 0 in its equation.

∴ 0 = 4 –

∴ = 4

∴ x = ± 2

∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)

Required area = area of the region APBOA

= 2[area of the region OPBO]

**3. Find the area of the region included between:**

**(i) = 2x and y = 2x.Solution:**

The vertex of the parabola = 2x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,

Area of the region OCBDO = area under the line y = 2x between x = 0 and x =

**(ii) = 4x and y = x.Solution:**
The vertex of the parabola = 4x is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,

Area of the region OCBDO = area under the line y = 2x between x = 0 and x =

**(iii) y = and the line y = 4x.Solution:**
The vertex of the parabola y = is at the origin 0(0, 0)

To find the points of the intersection of a line and the parabola.

Equating the values of y from the two equations, we get

= 4x

∴ – 4x = 0

∴ x(x – 4) = 0

∴ x = 0, x = 4

When x = 0, y = 4(0) = 0

When x = 4, y = 4(4) = 16

∴ the points of intersection are 0(0, 0) and B(4, 16)

Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)

Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4

**(iv) = 4ax and y = x.Solution:**

The vertex of the parabola = 4ax is at the origin O = (0, 0).

To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,

Area of the region OCBDO

= area under the line y

**(v) y = + 3 and y = x + 3.Solution:**

The given parabola is y = + 3, i.e. (x – 0 = y – 3

∴ its vertex is P(0, 3).

To find the points of intersection of the line and the parabola.

Equating the values of y from both the equations, we get

+ 3 = x + 3

∴ – x = 0

∴ x(x – 1) = 0

∴ x = 0 or x = 1

When x = 0, y = 0 + 3 = 3

When x = 1, y = 1 + 3 = 4

∴ the points of intersection are P(0, 3) and B(1, 4)

Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO

Now, area of the region OPABDO

= area under the line y = x + 3 between x = 0 and x = 1

Area of the region OPCBDO = area under the parabola y = + 3 between x = 0 and x = 1