Solve $\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx} $.
Answer is 1007.
I tried multiplying $\sqrt{x}-\sqrt{2014-x}\;$,
which results in $\frac{\sqrt{2014-x}(\sqrt{x}-\sqrt{2014-x})}{2x-2014}=$$\frac{\sqrt{2014x-x^2}}{2x-2014}-… \\ =\frac{\sqrt{2014/x-1}}{2-2014/x}-…
$
I got stuck so I tried substituting $u=2014-x$,
thus $\int_{0}^{2014}{\frac{u}{\sqrt{2014-u}+u}}du=… ?$
I found the value of 1007 using the value of integrand at x=0, 1007 and 2014.
But cannot solve integral. How can it be solved?
Best Answer
Let $I = \int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}$. Then, as you said, consider the substitution $u = 2014 - x$.
In particular,
$$I = \int_{2014}^0 \frac{\sqrt{u}}{\sqrt{2014 - u} + \sqrt{u}} \cdot (-1) \,du = \int_0^{2014} \frac{\sqrt{u}}{\sqrt{2014 - u} + \sqrt{u}} \,du$$
Relabel the $u$ as $x$, then, adding the integrals together, we get $$2I = \int_0^{2014} 1 \,dx$$ which gives you the result.