Match List I with List II

Choose the correct answer from the options given below :

A-III, B-II, C-IV, D-I

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