Question

The value of $\underset{2}{\overset{8}{\int }}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx$ is

• A. 10
• B. 0
• C. 8
• D. 3

Answer 1

Answer: (D)

Let $I=\underset{2}{\overset{8}{\int }}\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx\dots$(i)
$\Rightarrow I=\underset{2}{\overset{8}{\int }}\frac{\sqrt{10-(10-x)}}{\sqrt{10-x}+\sqrt{10-(10-x)}}dx$
$\left[\because \underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx\right]$
$\Rightarrow I=\underset{2}{\overset{8}{\int }}\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx\dots$(ii)
On adding Eqs. (i) and (ii), we get
$2I=\underset{2}{\overset{8}{\int }}\frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx$
$\Rightarrow 2I=\underset{2}{\overset{8}{\int }}1dx=[x{]}_2^8$
$\Rightarrow 2I=(8-2)=6$
$\therefore I=\frac{6}{2}=3$