If $f(x) = x^3 \sin^4 x$, then $\int_{-\pi/4}^{\pi/4} f(x)\,dx$ equals:
- $0$
- $\pi/4$
- $1$
- $2\int_0^{\pi/4} f\,dx$
Show answer and solution
Correct answer: A. $0$
$f(-x) = (-x)^3 \sin^4(-x) = -x^3 \cdot \sin^4 x = -f(x)$.
So $f$ is **odd**, and over the symmetric interval the integral is 0.