Area under a curve: the sign trap that loses 4 marks

Published 2026-04-01 · Written by Pete Bromfield

The definite integral gives you signed area. If part of your curve is below the x-axis, integration returns a negative value — and if you don't split, you'll be off by 4 marks. This is the drill.

The trap

Find the area between the curve $y = x^3 - 4x$ and the x-axis, from $x = -2$ to $x = 2$.

If you just compute $\int_{-2}^{2} (x^3 - 4x)\,dx$, you get zero — because the function is odd. But the area is not zero.

The fix — three steps

1. Find the roots: $x^3 - 4x = x(x-2)(x+2) = 0$, so $x = -2, 0, 2$.
2. Integrate over each region separately and take the absolute value: $\left|\int_{-2}^{0}(x^3 - 4x)\,dx\right| + \left|\int_{0}^{2}(x^3 - 4x)\,dx\right|$.
3. Each integral evaluates to $\pm 4$, so the area is $4 + 4 = 8$.

Mark-scheme note

The M-mark is for splitting the integral at the root. The A-mark is for the final $8$. If you skip the split, you get zero on both.

10 area-under-curve drills on Skills Practice → Integration.