alleryIf $f(x) = x^3 \sin^4 x$, then $\int_{-\pi/4}^{\pi/4} f(x)\,dx$ equals:
A$0$
B$\pi/4$
C$1$
D$2\int_0^{\pi/4} f\,dx$
Answer & 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.
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