JAMB 1997 · UME · Q42

If $y = x(x^4 + x^2 + 1)$ , evaluate $\displaystyle\int_{-1}^{1} y\,dx$ .

Worked solution (try it first)

Expand: $y = x^5 + x^3 + x$ , so the integral is $\left[\frac{x^6}{6} + \frac{x^4}{4} + \frac{x^2}{2}\right]_{-1}^{1}$ .
Every power is even, so the value at $x = 1$ and at $x = -1$ is the same, $\frac{11}{12}$ .
Subtract: $\frac{11}{12} - \frac{11}{12} = 0$ , option D.