Which of the following statements are correct?
(A) \(\int_{-1}^{1}\displaystyle\frac{x^{3}}{1 + x^{2}}\,dx = 0\)
(B) \(\int_{-1}^{1}\displaystyle x^{4}\,dx = 2\int_{0}^{1}\displaystyle x^{4}\,dx\)
(C) \(\int_{0}^{\frac{\pi}{2}}\displaystyle\sin^{4}x\,dx = \int_{0}^{\frac{\pi}{2}}\displaystyle\cos^{4}x\,dx\)
(D) \(\int_{-1}^{1}\displaystyle |x|\,dx = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
On \([-a, a]\): the integral of an odd function is 0, and the integral of an even function is twice the integral on \([0, a]\). On \([0, a]\): \(\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx\).
\( \displaystyle \frac{(-x)^{3}}{1 + (-x)^{2}} = -\frac{x^{3}}{1 + x^{2}} \) (odd)
\( (-x)^{4} = x^{4} \) (even)
\( |{-x}| = |x| \) (even)
\( \displaystyle \sin\left(\frac{\pi}{2} - x\right) = \cos x \)
\( \displaystyle \int_{-1}^{1}|x|\,dx = 2\int_{0}^{1}x\,dx = 1 \)
So \(\frac{x^{3}}{1 + x^{2}}\) is odd, \(x^{4}\) is even, replacing x by \(\frac{\pi}{2} - x\) turns \(\sin^{4}x\) into \(\cos^{4}x\), and \(\int_{-1}^{1}|x|\,dx = 1\).
Checking the statements
(A) \(\int_{-1}^{1}\frac{x^{3}}{1 + x^{2}}\,dx = 0\) — Correct. The integrand is odd.
(B) \(\int_{-1}^{1}x^{4}\,dx = 2\int_{0}^{1}x^{4}\,dx\) — Correct. The integrand is even.
(C) \(\int_{0}^{\frac{\pi}{2}}\sin^{4}x\,dx = \int_{0}^{\frac{\pi}{2}}\cos^{4}x\,dx\) — Correct. Replacing x by \(\frac{\pi}{2} - x\) changes \(\sin^{4}x\) to \(\cos^{4}x\).
(D) \(\int_{-1}^{1}|x|\,dx = 0\) — Incorrect. |x| is even and never negative; the value is 1.