Which of the following integrals are equal to zero?
(A) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle x^{3}\cos x\,dx\)
(B) \(\int_{0}^{\pi}\displaystyle\cos^{3}x\,dx\)
(C) \(\int_{-1}^{1}\displaystyle x^{2}|x|\,dx\)
(D) \(\int_{0}^{\pi}\displaystyle\sin^{3}x\,dx\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
\(x^{3}\cos x\) is an odd function, since \((-x)^{3}\cos(-x) = -x^{3}\cos x\), so its integral over \([-\frac{\pi}{2}, \frac{\pi}{2}]\) is 0.
For (B), let \(I = \int_{0}^{\pi}\cos^{3}x\,dx\). Replacing x by \(\pi - x\) and using \(\cos(\pi - x) = -\cos x\):
\( \displaystyle I = \int_{0}^{\pi}(-\cos x)^{3}\,dx = -I \)
\(2I = 0,\quad I = 0\)
For (C), \(x^{2}|x|\) is even and never negative:
\( \displaystyle \int_{-1}^{1}x^{2}|x|\,dx = 2\int_{0}^{1}x^{3}\,dx = \frac{1}{2} \)
For (D), using \(\sin^{3}x = \sin x - \sin x\cos^{2}x\):
\( \displaystyle \int_{0}^{\pi}\sin^{3}x\,dx = \left[-\cos x + \frac{\cos^{3}x}{3}\right]_{0}^{\pi} \)
\( \displaystyle = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{4}{3} \)
So the values are (A) 0, (B) 0, (C) \(\frac{1}{2}\) and (D) \(\frac{4}{3}\).
Checking the statements
(A) \(\int_{-\pi/2}^{\pi/2}x^{3}\cos x\,dx\) — Correct. Odd integrand on a symmetric interval, value 0.
(B) \(\int_{0}^{\pi}\cos^{3}x\,dx\) — Correct. Replacing x by \(\pi - x\) gives \(I = -I\), so the value is 0.
(C) \(\int_{-1}^{1}x^{2}|x|\,dx\) — Incorrect. The integrand is even and positive; the value is \(\frac{1}{2}\).
(D) \(\int_{0}^{\pi}\sin^{3}x\,dx\) — Incorrect. \(\sin^{3}x \gt 0\) on \((0, \pi)\); the value is \(\frac{4}{3}\).