# Thread: very hard integration question

1. ## very hard integration question

I need some help here, this is a hard one. Show me some steps.

For the following integral
$\displaystyle I = \int^{10}_{0}\frac{\sqrt{x}}{\sqrt{x}+{\sqrt{10-x}}}$
use a substitution to show that
$\displaystyle I = \int^{10}_{0}\frac{\sqrt{10-x}}{\sqrt{x}+{\sqrt{10-x}}}$
Use these two representations of I to evaluate I.

2. Originally Posted by gammaman
I need some help here, this is a hard one. Show me some steps.

For the following integral
$\displaystyle I = \int^{10}_{0}\frac{\sqrt{x}}{\sqrt{x}+{\sqrt{10-x}}} \, {\color{red}dx}$
use a substitution to show that
$\displaystyle I = \int^{10}_{0}\frac{\sqrt{10-x}}{\sqrt{x}+{\sqrt{10-x}}} \, {\color{red}dx}$
Use these two representations of I to evaluate I.
First note what I've added in red. Sloppy notation (such as your omissions) always leads to difficulty and trouble and lost marks.

Substitute $\displaystyle x = 10 - u$ to get $\displaystyle I = \int^{10}_{0}\frac{\sqrt{10-u}}{\sqrt{u}+{\sqrt{10-u}}} \, du$.

Note that $\displaystyle u$ is just a dummy variable in the definite integral and can be therefore replaced with any symbol, including x ....

Now add the two expressions for I to get $\displaystyle 2I = \int_0^{10} dx \, ....$