Match List-I with List-II.
| List-I | List-II | ||
| (A) | \(\int_{-1}^{1}\displaystyle |x|\,dx\) | (I) | 0 |
| (B) | \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle \sin x\,dx\) | (II) | 2 |
| (C) | \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle \cos x\,dx\) | (III) | \(\displaystyle\frac{\pi}{4}\) |
| (D) | \(\int_{0}^{\frac{\pi}{2}}\displaystyle \sin^2 x\,dx\) | (IV) | 1 |
Choose the correct option from the ones given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(A) \(\int_{-1}^{1}|x|\,dx\)
This matches (IV) 1.
|x| is an even function, so \(\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx\):
\( \displaystyle \int_{-1}^{1}|x|\,dx = 2\int_{0}^{1}x\,dx = 2\left[\frac{x^2}{2}\right]_{0}^{1} = 1 \)
(B) \(\int_{-\pi/2}^{\pi/2}\sin x\,dx\)
This matches (I) 0.
sin x is an odd function, and the integral of an odd function over [−a, a] is 0:
\( \displaystyle \int_{-\pi/2}^{\pi/2}\sin x\,dx = 0 \)
(C) \(\int_{-\pi/2}^{\pi/2}\cos x\,dx\)
This matches (II) 2.
cos x is even:
\( \displaystyle \int_{-\pi/2}^{\pi/2}\cos x\,dx = 2\int_{0}^{\pi/2}\cos x\,dx = 2\left[\sin x\right]_{0}^{\pi/2} = 2 \)
(D) \(\int_{0}^{\pi/2}\sin^2 x\,dx\)
This matches (III) \(\frac{\pi}{4}\).
Using \(\sin^2 x = \frac{1 - \cos 2x}{2}\):
\( \displaystyle \int_{0}^{\pi/2}\sin^2 x\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_{0}^{\pi/2} \)
\( \displaystyle = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)