Chapter 8: Application of Integrals

Given curve y = √(x – 1)

⇒ y² = x – 1

Plotting the curve and finding the area of the shaded region between the lines x = 1 and x = 5, we have

Therefore, the area of the required region = 16/3 sq.units

Given curve y = √(a² – x²) and lines x = 0 and x = a

y = √(a² – x²) ⇒ y² = a² – x²

x² + y² = a² which is equation of a circle.

Now, the required region is found by plotting the curve and lines.

So, the area of the shaded region is

Therefore, the required area = πa²/4

Given equations of curve y = √x and line y = x

Solving the equations y = √x ⇒ y² = x and y = x, we get

x² = x

x² – x = 0

x(x – 1) = 0

So, x = 0, 1

Now, the required area of the shaded region

Therefore, the required area = 1/6 sq.units.

Given curve y = –x² or x² = –y and the line x + y + 2 = 0

Solving the two equation, we get

x – x² + 2 = 0

x² – x – 2 = 0

x² – 2x + x – 2 = 0

x(x – 2) + 1 (x – 2) = 0

(x – 2) (x + 1) = 0

So, x = -1, 2

Now,

The area of the required shaded region

Therefore, the required area = 9/2 sq.units

Given curve y = √x and line x = 2y + 3, first quadrant and x-axis.

Solving y = √x and x = 2y + 3, we get

Long Answer (L.A.)

Given equation of curves are y² = 2x and x² + y² = 4x

Solving the equations, we have

x² – 4x + y² = 0

x² – 4x + 4 – 4 + y² = 0

(x – 2)² + y² = 4

It’s clearly seen that the equation of the circle having its centre (2, 0) and radius 2.

Solving x² + y² = 4x and y² = 2x

x² + 2x = 4x

x² + 2x – 4x = 0

x² – 2x = 0

x(x – 2) = 0

So, x = 0, 2

Now, the area of the required region is given as