Question

Find the area enclosed between 3y = x², X-axis and x = 2 to x = 3.

Answer 1

3y = x²

∴ y = `x^2/3`

Required area = `int_2^3 y  dx`

= `int_2^3 x^2/3dx`

= `[x^3/9]_2^3`

= `3 - 8/9`

= `(27 - 8)/9`

= `19/9` sq.units