Question

Draw a rough sketch of the curves y² = x and y² = 4 – 3x and find the area enclosed between them.

Answer 1

Given equation of curves are,

y² = x   ...(i)

And y² = 4 – 3x   ...(ii)

These are the equations of parabola.

Solving equations (i) and (ii), we get

x = 4 – 3x

⇒ 4x = 4

⇒ x = 1

∴ y² = 1

⇒ y = ±1

So, (1, 1) and (1, –1) are the points of intersection of the two parabolas.

∴ Required area = `int_(-1)^1 x_"II"  dy - int_(-1)^1 x_"I" dy`

= `int_(-1)^1 (4 - y^2)/3 dy - int_(-1)^1 y^2 dy`   ...`[(x_"II" = (4  -  y^2)/3),(x_"I" = y^2)]`

= `1/3 [4y - y^3/3]_(-1)^1 - [y^3/3]_(-1)^1`

= `1/3 [(4 xx 1 - 1^3/3) - (4 xx (-1) - (-1)^3/3)] - [1^3/3 - (-1)^3/3]`

= `1/3 [4 - 1/3 + 4 - 1/3] - [1/3 + 1/3]`

= `1/3 [8 - 2/3] - 2/3`

= `1/3 xx 22/3 - 2/3`

= `22/9 - 2/3`

= `(22 - 6)/9`

= `16/9` sq. units.