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

Question

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.

Try the substitution $u=2012-x$ – Display name Dec 18 '14 at 09:13 — Display name, Dec 18 '14 at 09:13

score 14 · Accepted Answer · answered Dec 18 '14 at 09:14

14

HINT:

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

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

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

answered Dec 18 '14 at 09:14

lab bhattacharjee

274,582

How does this property suffice? – Amad27 Dec 18 '14 at 09:15
@Amad27, if $f(x)=\dfrac{\sqrt x }{\sqrt{2012 - x} + \sqrt x}, f(1+2011-x)=?$ – lab bhattacharjee Dec 18 '14 at 09:16
$f(2012 - x) = \frac{\sqrt{2012 - x}}{\sqrt{x} + \sqrt{2012 - x}}$ – Amad27 Dec 18 '14 at 09:18
@Amad27, $f(x)+f(1+2011-x)=?$ – lab bhattacharjee Dec 18 '14 at 09:18
$f(x) + f(2012 -x ) = \frac{\sqrt{x} + \sqrt{2012 - x}}{\sqrt{x} + \sqrt{2012 - x}} = 1$ So the integral answer is 2011 - 1 = 2010? – Amad27 Dec 18 '14 at 09:20
Divided by 2 = 1005 – Amad27 Dec 18 '14 at 09:21
@Amad27, Yes. See also : http://math.stackexchange.com/questions/578957/definite-integral-int-24-frac-sqrt-log9-x-sqrt-log9-x-sqrt-log3 and http://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx – lab bhattacharjee Dec 18 '14 at 09:21
Let us continue this discussion in chat. – lab bhattacharjee Dec 18 '14 at 09:24

score 6 · Answer 2 · edited Dec 03 '19 at 20:33

6

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

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

edited Dec 03 '19 at 20:33

Andrés Villa

842

answered Dec 18 '14 at 09:16

Display name

611

Hello friend, you have a small error, the integration limits are $ u (1) $, $ u(2011) $ – Andrés Villa Dec 03 '19 at 19:48

score 0 · Answer 3 · answered Dec 20 '22 at 21:16

0

It is known that : $$\int_a^b\frac{f(x)}{f(a+b-x)+f(x)}dx=\frac{a+b}{2}$$You should remember this. After making sure that this is the case, you get the answer as $1005$.

answered Dec 20 '22 at 21:16

Kamal Saleh

6,497

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

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