Chapter 8: Application of Integrals

Given curves are y² = 9x and y = 3x

Now, solving the two equations we have

(3x)² = 9x

9x² = 9x

9x² – 9x = 0 ⇒ 9x(x – 1) = 0

Thus, x = 0, 1

So, the area of the shaded region is given by

= ar(region OAB) – ar (Δ OAB)

Therefore, the required area is ½ sq.units.

Given parabolas are y² = 2px …. (i) and x² = 2py …. (ii)

Now, from equation (ii) we have

y = x²/2p

Putting the value of y in equation (i), we have

(x²/2p)² = 2px

x⁴/4p² = 2px

x⁴ = 8p³x

x⁴ – 8p³x = 0

x(x³ – 8p³) = 0

So, x = 0 or x³ – 8p³ = 0 ⇒ x = 2p

Now, the required area is

Therefore, the required area is 4/3 p² sq. units.

Given curves are y = x³, y = x + 6 and x = 0

On solving y = x³ and y = x + 6, we have

x³ = x + 6

x³ – x – 6 = 0

x²(x – 2) + 2x(x – 2) + 3(x – 2) = 0

(x – 2) (x² + 2x + 3) = 0

It’s seen that x² + 2x + 3 = 0 has no real roots

So, x = 2 is the only root for the above equation.

Now, the required area of the shaded region is given by

Given curves are y² = 4x … (i) and x² = 4y … (ii)

On solving the equations, we get

From (ii),

y = x²/4

Putting value of y in (i), we have

(x²/4)² = 4x

x⁴/16 = 4x

x⁴ = 64x

x⁴ – 64x = 0

So, x = 0, 4

Now, the required area is the shaded region

Given curves are y² = 9x and y = x

Solving the above equations, we have

x² = 9x ⇒ x² – 9x = 0

x(x – 9) = 0

So, x = 0, 9

Now, the required area is

Therefore, the required area is 27/2 sq. units.

Given, equation of parabola x² = y and line y = x + 2

Solving the above equations, we get

x² = x + 2

x² – x – 2 = 0

x² – 2x + x – 2 = 0

x(x – 2) + 1(x – 2) = 0

(x + 1) (x – 2) = 0

So, x = -1, 2

Now,

Therefore, the area of the required region is 9/2 sq. units.

Given, equation of line x = 2 and parabola y² = 8x

Putting value of x in the other equation, we have

y² = 8(2)

y² = 16

So, y = ± 4

Now, the required area is

Therefore, the area of the region = 32/3 sq. units

Given, {(x, 0) : y = √(4 – x²)}

So, y² = 4 – x²

x² + y² = 4 which is a circle.

Now, the required area

Given, 2y = 5x + 7, x-axis, x = 2 and x = 8

Let’s draw the graph of 2y = 5x + 7 ⇒ y = (5x + 7)/2

Now, let’s plot the straight line on a graph with other lines.

The area of the required region is

Therefore, the required area the region = 96 sq. units