Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \int_{-1}^{1}\displaystyle|x|dx \) | (I) | 0 |
| (B) | \( \int_{-1}^{1}\displaystyle xdx+2 \) | (II) | \( \displaystyle\frac{1}{2} \) |
| (C) | \( \int_{-1}^{1}\displaystyle\sin^5x\cos^4x\,dx \) | (III) | 1 |
| (D) | \( \int_{0}^{\frac{\pi}{4}}\displaystyle\sin 2x\,dx \) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Over a symmetric interval [−a, a], an even function gives \( 2\int_{0}^{a}f(x)\,dx \) and an odd function gives 0.
(A) \( \int_{-1}^{1}|x|\,dx \)
This matches (III) 1. |x| is even:
\( \displaystyle \int_{-1}^{1}|x|\,dx=2\int_{0}^{1}x\,dx \)
\( \displaystyle =2\times\frac{1}{2}=1 \)
(B) \( \int_{-1}^{1}x\,dx+2 \)
This matches (IV) 2. x is odd, so the integral is 0:
\( \displaystyle 0+2=2 \)
(C) \( \int_{-1}^{1}\sin^5x\cos^4x\,dx \)
This matches (I) 0. With \( f(x)=\sin^5x\cos^4x \), \( f(-x)=(-\sin x)^5\cos^4x=-f(x) \), so f is odd and the integral is 0.
(D) \( \int_{0}^{\frac{\pi}{4}}\sin 2x\,dx \)
This matches (II) \( \frac{1}{2} \).
\( \displaystyle \int_{0}^{\pi/4}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_{0}^{\pi/4} \)
\( \displaystyle =-\frac{\cos\frac{\pi}{2}}{2}+\frac{\cos 0}{2} \)
\( \displaystyle =0+\frac{1}{2}=\frac{1}{2} \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).