WAEC 2020 · Paper 1 · Q4

If $\displaystyle\int_0^3 (px^2 + 16)\,dx = 129$ , find the value of $p$ .

Worked solution (try it first)

Integrate term by term: $\displaystyle\int (px^2 + 16)\,dx = \frac{px^3}{3} + 16x$ .
Put in the limits: $\left(\frac{27p}{3} + 48\right) - 0 = 9p + 48$ .
Set $9p + 48 = 129$ , so $9p = 81$ .
So $p = 9$ , option A.