Definite Integration Question 71
Question: The value of the integral $ \int _{\frac{1}{n}}^{\frac{an-1}{n}}{\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx} $ is
[AMU 2002]
Options:
A) $ \frac{a}{2} $
B) $ \frac{na+2}{2n} $
C) $ \frac{na-2}{2n} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int _{1/n}^{\frac{an-1}{n}}{\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx=\int _{1/n}^{a-\frac{1}{n}}{\frac{\sqrt{x}}{\sqrt{a-x}+\sqrt{x}}dx}} $ …..(i)
$ =\int _{\frac{1}{n}}^{a-\frac{1}{n}}{\frac{\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}dx}{\sqrt{a-( \frac{1}{n}+a-\frac{1}{n}-x )+}\sqrt{\frac{1}{n}+a-\frac{1}{n}-x}}} $
$ $
$ [ \because \int_a^{b}{f(x)dx=\int_a^{b}{f(a+b-x)dx}} ] $
$ I=\int _{\frac{1}{n}}^{a-\frac{1}{n}}{\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx} $ ……(ii)
Adding (i) and (ii), we get $ 2I=\int _{1/n}^{a-(1/n)}{1dx=[ x ] _{1/n}^{a-\frac{1}{n}}} $
$ \Rightarrow 2I=a-\frac{1}{n}-\frac{1}{n}=\frac{na-2}{n} $
$ \Rightarrow I=\frac{na-2}{2n} $ .
BETA
