Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \int_{-1}^{1}|x|\,dx= \) | (I) | 3 |
| (B) | \( \int_{-\pi}^{\pi}\cos x\,dx= \) | (II) | 1 |
| (C) | \( \int_{-1}^{1}(|x|-1)\,dx= \) | (III) | 0 |
| (D) | \( \int_{-1}^{1}(|x|+1)\,dx= \) | (IV) | -1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
(A) \( \int_{-1}^{1}|x|\,dx \)
This matches (II) 1. \( |x| \) is an even function, so
\( \displaystyle \int_{-1}^{1}|x|\,dx=2\int_{0}^{1}x\,dx \)
\( \displaystyle =2\left[\frac{x^2}{2}\right]_{0}^{1} \)
\( \displaystyle =1 \)
(B) \( \int_{-\pi}^{\pi}\cos x\,dx \)
This matches (III) 0.
\( \displaystyle \int_{-\pi}^{\pi}\cos x\,dx=\Big[\sin x\Big]_{-\pi}^{\pi} \)
\( \displaystyle =\sin\pi-\sin(-\pi) \)
\( \displaystyle =0 \)
(C) \( \int_{-1}^{1}(|x|-1)\,dx \)
This matches (IV) -1. Splitting the integral and using the value 1 from (A),
\( \displaystyle \int_{-1}^{1}(|x|-1)\,dx=\int_{-1}^{1}|x|\,dx-\int_{-1}^{1}1\,dx \)
\( \displaystyle =1-2=-1 \)
(D) \( \int_{-1}^{1}(|x|+1)\,dx \)
This matches (I) 3.
\( \displaystyle \int_{-1}^{1}(|x|+1)\,dx=\int_{-1}^{1}|x|\,dx+\int_{-1}^{1}1\,dx \)
\( \displaystyle =1+2=3 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
Shortcut: (A) is the area of two right triangles with legs 1 and 1, which is 1. Once (A) is known, (C) and (D) are 1 − 2 and 1 + 2, since the interval [−1, 1] has length 2.