Match List-I with List-II.
| List-I | List-II | ||
| (A) | $\int\limits^{1}_{-1}|x|dx$ | (I) | 0 |
| (B) | $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}sinx\, dx$ | (II) | 2 |
| (C) | $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}cos x\, dx$ | (III) | $\frac{\pi }{4}$ |
| (D) | $\int\limits^{\frac{\pi }{2}}_{0}sin^2x\, dx$ | (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) $\int\limits^{1}_{-1}|x|dx=2\int\limits_0^1x\,dx=1$ (IV)
(B) $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}\sin x\, dx=0$ (Odd function) (I)
(C) $\int\limits^{\frac{\pi }{2}}_{-\frac{\pi }{2}}\cos x\, dx=\left[\sin x\right]^{\frac{\pi }{2}}_{-\frac{\pi }{2}}=2$ (II)
(D) $\int\limits^{\frac{\pi }{2}}_{0}\sin^2x\, dx=\int\limits^{\frac{\pi }{2}}_{0}\frac{1-\cos 2x}{2}dx=\frac{\pi}{4}$ (III)