Match List-I with List-II
| List-I | List-II | ||
| Definite integral | Value | ||
| (A) | \( \int_{2}^{3}\displaystyle x^{2}\,dx \) | (I) | \( -\displaystyle\frac{1}{2} \) |
| (B) | \( \int_{-1}^{1}\displaystyle (x+1)\,dx \) | (II) | \( \displaystyle\frac{19}{3} \) |
| (C) | \( \int_{0}^{1}\displaystyle (x-1)\,dx \) | (III) | \( \displaystyle\frac{16}{3} \) |
| (D) | \( \int_{0}^{4}\displaystyle \sqrt{x}\,dx \) | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
(A) \( \int_{2}^{3}x^{2}\,dx \)
This matches (II) \( \frac{19}{3} \).
\( \displaystyle \left[\frac{x^3}{3}\right]_2^3=\frac{27-8}{3} \)
\( \displaystyle =\frac{19}{3} \)
(B) \( \int_{-1}^{1}(x+1)\,dx \)
This matches (IV) 2.
\( \displaystyle \left[\frac{x^2}{2}+x\right]_{-1}^{1}=\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right) \)
\( \displaystyle =2 \)
(C) \( \int_{0}^{1}(x-1)\,dx \)
This matches (I) \( -\frac{1}{2} \).
\( \displaystyle \left[\frac{x^2}{2}-x\right]_0^1=\frac{1}{2}-1 \)
\( \displaystyle =-\frac{1}{2} \)
(D) \( \int_{0}^{4}\sqrt{x}\,dx \)
This matches (III) \( \frac{16}{3} \).
\( \displaystyle \left[\frac{2}{3}x^{3/2}\right]_0^4=\frac{2}{3}\times 8 \)
\( \displaystyle =\frac{16}{3} \)
So the matching is (A) - (II), (B) - (IV), (C) - (I), (D) - (III).
Shortcut: (C) is the only integral that is negative, since x − 1 ≤ 0 on [0, 1]; so (C) - (I). Only Option (4) has it.