Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \int_{0}^{1}(x^{3}+x)\,dx \) | (I) | 0 |
| (B) | \( \int_{-1}^{1}(x^{3}+1)\,dx \) | (II) | \( \displaystyle\frac{5}{4} \) |
| (C) | \( \int_{-1}^{1}(x^{3}+x)\,dx \) | (III) | 2 |
| (D) | \( \int_{0}^{1}(x^{3}+1)\,dx \) | (IV) | \( \displaystyle\frac{3}{4} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
(A) \( \int_{0}^{1}(x^{3}+x)\,dx \)
This matches (IV).
\( \displaystyle \int_{0}^{1}(x^{3}+x)\,dx=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} \)
(B) \( \int_{-1}^{1}(x^{3}+1)\,dx \)
This matches (III).
The function \( x^{3} \) is odd, so its integral over \( [-1,1] \) is 0. What remains is \( \int_{-1}^{1}1\,dx=2 \).
(C) \( \int_{-1}^{1}(x^{3}+x)\,dx \)
This matches (I).
The function \( x^{3}+x \) is odd, so its integral over \( [-1,1] \) is 0.
(D) \( \int_{0}^{1}(x^{3}+1)\,dx \)
This matches (II).
\( \displaystyle \int_{0}^{1}(x^{3}+1)\,dx=\frac{1}{4}+1=\frac{5}{4} \)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).