WAEC Further Mathematics 2022 Paper

$\int\limits_{0}^{1} x^2(x^3+2)^3$dx

let $u = x^3 + 2, du = 3x^2dx$

when x = 1, u = 3

when x = 0, u = 2

dx = $\;\frac{du}{3x^2}$

$\int\limits_{2}^{3}$ $\;\frac{x^2[u]^3}{3x^2}$

$\int\limits_{2}^{3}$ $\;\frac{u^3}{3}$ du

= $\;\frac{u^4}{3\cdot4}$$_2$3

$\;\frac{1}{12} [u^4]$$_2$3

$\;\frac{1}{12} [3^4 - 2^4]$

$\;\frac{1}{12}[81 - 16]$

$\;\frac{65}{12}$