Match List I with List II
| LIST I | LIST II | ||
| A. | $\int\limits^{2}_{1}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$ | I. | $-\frac{1}{2}log3$ |
| B. | $\int\limits^{2}_{1}x^2dx$ | II. | $\frac{7}{3}$ |
| C. | $\int\limits^{-1}_{-4}\frac{1}{x}dx$ | III. | $\frac{1}{2}$ |
| D. | $\int\limits^{1}_{0}\frac{1}{2x-3}dx$ | IV. | $-log4$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
Evaluate each integral:
A. $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx$
Put $x=3-t$ implies $dx=-dt$. Changing limits: $x=1\to t=2$, $x=2\to t=1$.
$I=\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx = \int_{1}^{2} \frac{\sqrt{3-t}}{\sqrt{t}+\sqrt{3-t}}dt$
Add both forms:
$2I=\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}dx + \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}dx = \int_{1}^{2} 1dx=1$
So $I=\frac{1}{2}$.
Match: A → III
B. $\int_{1}^{2} x^2 dx = \left[\frac{x^3}{3}\right]_{1}^{2} = \frac{8}{3}-\frac{1}{3}=\frac{7}{3}$
Match: B → II
C. $\int_{-4}^{-1}\frac{1}{x}dx = [\log|x|]_{-4}^{-1} = \log(1)-\log(4) = -\log4$
Match: C → IV
D. $\int_{0}^{1}\frac{1}{2x-3}dx$
Let $u=2x-3$ implies $dx=\frac{du}{2}$
Limits: $x=0$ implies $u=-3$, $x=1$ implies $u=-1$
$I=\frac{1}{2}\int_{-3}^{-1}\frac{1}{u}du=\frac{1}{2}[\log|u|]_{-3}^{-1}=\frac{1}{2}(\log1-\log3)=-\frac{1}{2}\log3$
Match: D → I