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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?

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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.

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  • $\begingroup$ Why can't I see it! Thank you. $\endgroup$ – nik Sep 7 '18 at 8:54
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If you substitute $x \mapsto 2014-x$, you get $$ I = \int_0^{2014} \frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} \, dx = \int_{0}^{2014} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx. $$ Hence \begin{align} 1007 &= \frac{1}{2}\int_0^{2014} dx = \frac{1}{2}\int_0^{2014}\frac{\sqrt{2014-x}}{\sqrt{x} + \sqrt{2014-x}} + \frac{\sqrt{x}}{\sqrt{x} + \sqrt{2014-x}} \, dx\\ &= \frac{1}{2}(I+I) = I. \end{align}

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$$let \int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}=A \\ put \ 2014-x=u \\ \implies \int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}=\int_{2014}^{0}{\frac{\sqrt{u}}{\sqrt{u}+\sqrt{2014-u}}(-du)}\\=\int_{0}^{2014}{\frac{\sqrt{u}}{\sqrt{u}+\sqrt{2014-u}}(du)}=A \\ add \ both \ integrals \ \implies 2(A)=\int_0^{2014}dx \implies A=1007$$

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Don't need to do that substitution again and again just apply this formula

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

$$I=\int_{0}^{2014}{\frac{\sqrt{2014-x}}{\sqrt{x}+\sqrt{2014-x}}dx}$$

Apply the formula

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

$$ I=\int_{0}^{2014}{\frac{\sqrt{2014-(2014+0-x)}}{\sqrt{2014+0-x}+\sqrt{2014-(2014+0-x)}}dx}=\int_{0}^{2014}{\frac{\sqrt x}{\sqrt{2014-x}+\sqrt{x}}dx}$$

add both integrals

$$2I=\int_{0}^{2014}dx$$ $$2I=2014$$ $$I=1007$$

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