WAEC 2017 · Paper 2 · Q4

Given that $y = 3x^2 - \dfrac{3}{x^2}$ , $x \ne 0$ , find:

Worked solution (try it first)

Integrate: $\displaystyle\int (3x^2 - 3x^{-2})\,dx = x^3 + 3x^{-1}$
$= x^3 + \dfrac3x$ .
At $x = 3$ : $27 + 1 = 28$ .
At $x = 1$ : $1 + 3 = 4$ .
So $\displaystyle\int_1^3 y\,dx = 28 - 4 = 24$ .