Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \int_{-1}^{1}(2x^5+3x^3+4x+1)\,dx= \) | (I) | 2 |
| (B) | \( \int_{-2}^{2}(4x^3+2)\,dx= \) | (II) | 12 |
| (C) | \( \int_{-1}^{1}x^3\,dx= \) | (III) | 8 |
| (D) | \( \int_{-3}^{3}(3x^5+2)\,dx= \) | (IV) | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (I), (B) - (III), (C) - (IV), (D) - (II)
On a symmetric interval [−a, a], odd powers of x integrate to 0, and a constant c integrates to 2ac.
(A) \( \int_{-1}^{1}(2x^5+3x^3+4x+1)\,dx \)
This matches (I) 2.
\( 2x^5 \), \( 3x^3 \) and 4x are odd, so only the constant 1 contributes:
\( \displaystyle \int_{-1}^{1}1\,dx=2 \)
(B) \( \int_{-2}^{2}(4x^3+2)\,dx \)
This matches (III) 8.
\( \displaystyle \int_{-2}^{2}2\,dx=2\times 4=8 \)
(C) \( \int_{-1}^{1}x^3\,dx \)
This matches (IV) 0.
\( x^3 \) is odd, so the integral is 0.
(D) \( \int_{-3}^{3}(3x^5+2)\,dx \)
This matches (II) 12.
\( \displaystyle \int_{-3}^{3}2\,dx=2\times 6=12 \)
So the matching is (A)-(I), (B)-(III), (C)-(IV), (D)-(II).