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Evaluate:

$$\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$

Using real methods only.

I am not sure what to do.

I tried finding a power series, which was too ugly.

I just need some hints, not an answer to do this integral, this is from the MIT Integration bee 2012.

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    $\begingroup$ Try the substitution $u=2012-x$ $\endgroup$ – Display name Dec 18 '14 at 9:13
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HINT:

As $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

So, if $\int_a^bf(x)\ dx=I,$

$$2I=\int_a^b[f(x)+f(a+b-x)]\ dx$$

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  • $\begingroup$ How does this property suffice? $\endgroup$ – Amad27 Dec 18 '14 at 9:15
  • $\begingroup$ @Amad27, if $f(x)=\dfrac{\sqrt x }{\sqrt{2012 - x} + \sqrt x}, f(1+2011-x)=?$ $\endgroup$ – lab bhattacharjee Dec 18 '14 at 9:16
  • $\begingroup$ $f(2012 - x) = \frac{\sqrt{2012 - x}}{\sqrt{x} + \sqrt{2012 - x}}$ $\endgroup$ – Amad27 Dec 18 '14 at 9:18
  • $\begingroup$ @Amad27, $f(x)+f(1+2011-x)=?$ $\endgroup$ – lab bhattacharjee Dec 18 '14 at 9:18
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    $\begingroup$ This is a great technique. Is there a proof of that property though? $\endgroup$ – Amad27 Dec 18 '14 at 9:21
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$$I=\int_{1}^{2011} \frac{\sqrt{x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$

Let $u=2012-x$ then $$I=\int_{1}^{2011} \frac{\sqrt{2012-u}}{\sqrt{2012 - u} + \sqrt{u}}du=\int_{1}^{2011} \frac{\sqrt{2012-x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$ Thus $$2I=\int_{1}^{2011} \frac{\sqrt{x}+\sqrt{2012 - x}}{\sqrt{2012 - x} + \sqrt{x}}dx$$

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