Question

The value of ${\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx$ is

• A. $10$
• B. $0$
• C. $8$
• D. $3$

Answer 1

The correct option is D $3$

Let $I={\int }_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}dx$ ..........(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_2^8\frac{\sqrt{10-(8+2-x)}}{\sqrt{10-x}+\sqrt{10-(8+2-x)}}dx$

$I={\int }_2^8\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}dx$ ........(2)

Add (1) and (2)

$2I={\int }_2^{8}dx=[x{]}_2^8=8-2=6$

$\therefore I=3$