Question

Evaluate the following integral:

\[\int_2^8 \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}dx\]

Answer 1

\[\text{Let I} =\int_2^8 \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}dx................(1)\]

Then,

\[I = \int_2^8 \frac{\sqrt{10 - \left( 2 + 8 - x \right)}}{\sqrt{2 + 8 - x} + \sqrt{10 - \left( 2 + 8 - x \right)}}dx .....................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_2^8 \frac{\sqrt{x}}{\sqrt{10 - x} + \sqrt{x}}dx ................(2)\]

Adding (1) and (2), we have

\[2I = \int_2^8 \frac{\sqrt{10 - x} + \sqrt{x}}{\sqrt{x} + \sqrt{10 - x}}dx\]
\[ \Rightarrow 2I = \int_2^8 dx\]
\[ \Rightarrow 2I = \left.x\right|_2^8 \]
\[ \Rightarrow 2I = 8 - 2 = 6\]
\[ \Rightarrow I = 3\]