Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (Integral) | (Value) | ||
| (A) | \(\int_{-1}^{1}\displaystyle (|x|+1)dx\) | (I) | 0 |
| (B) | \(\int_{-2}^{2}\displaystyle |x+1|dx\) | (II) | 2 |
| (C) | \(\int_{-1}^{1}\displaystyle 3|x|^2dx\) | (III) | 5 |
| (D) | \(\int_{-1}^{1}\displaystyle x|x|dx\) | (IV) | 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
On a symmetric interval [−a, a], an even function gives \(2\int_0^a f(x)\,dx\) and an odd function gives 0.
(A) \(\int_{-1}^{1} (|x|+1)\,dx\)
This matches (IV) 3. The integrand is even:
\(\displaystyle \int_{-1}^{1} (|x|+1)\,dx = 2\int_{0}^{1} (x + 1)\,dx = 2\left[\frac{x^2}{2} + x\right]_{0}^{1}\)
\(\displaystyle = 2 \times \frac{3}{2} = 3\)
(B) \(\int_{-2}^{2} |x+1|\,dx\)
This matches (III) 5. Since \(x + 1\) changes sign at x = −1, split there:
\(\displaystyle \int_{-2}^{-1} -(x+1)\,dx = \left[-\frac{(x+1)^2}{2}\right]_{-2}^{-1} = 0 + \frac{1}{2} = \frac{1}{2}\)
\(\displaystyle \int_{-1}^{2} (x+1)\,dx = \left[\frac{(x+1)^2}{2}\right]_{-1}^{2} = \frac{9}{2} - 0 = \frac{9}{2}\)
\(\displaystyle \frac{1}{2} + \frac{9}{2} = 5\)
(C) \(\int_{-1}^{1} 3|x|^2\,dx\)
This matches (II) 2, since \(|x|^2 = x^2\):
\(\displaystyle \int_{-1}^{1} 3x^2\,dx = \left[x^3\right]_{-1}^{1} = 1 - (-1) = 2\)
(D) \(\int_{-1}^{1} x|x|\,dx\)
This matches (I) 0. The integrand is odd: \((-x)|-x| = -x|x|\).
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).