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Evaluate $\displaystyle\int_{0}^{2} \int_{0}^{\log(x)}(x-1)\sqrt{1+e^y}\,dy\,dx$. I have tried integration by substitution but can't connect to type 1 or type 2. Any help.

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  • $\begingroup$ Is the order of integration $dxdy$ or $dydx$? $\endgroup$
    – John_dydx
    Jul 10, 2015 at 19:02
  • $\begingroup$ @john it's dydx $\endgroup$ Jul 10, 2015 at 19:04

2 Answers 2

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Change the order of integration to obtain

$$\begin{align} I&\equiv\int_0^1\int_0^{\log x}(x-1)\sqrt{1+e^y}dydx\\\\&=-\int_{-\infty}^0\int_0^{e^y}(x-1)\sqrt{1+e^y}dxdy\\\\&=-\int_{-\infty}^0\left(\frac12 e^{2y}-e^y\right)\sqrt{1+e^y}dxdy \end{align}$$

Then, letting $y\to \log y$, we have

$$\begin{align} I&=-\int_0^1\left(\frac12 y-1\right)\sqrt{1+y}dy\\\\ &=-\int_1^2\left(\frac12 y-\frac32\right)y^{1/2}dy\\\\ &=\frac25\left(3\sqrt{2}-2\right) \end{align}$$

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  • $\begingroup$ Nice answer! I was sort of thinking along the same lines. $\endgroup$
    – John_dydx
    Jul 10, 2015 at 21:04
  • $\begingroup$ Thanks @John! I'm at home and recovering from surgery last week and you just made my day! $\endgroup$
    – Mark Viola
    Jul 10, 2015 at 21:07
  • $\begingroup$ Get well soon Dr. MV! Enjoy the weekend. $\endgroup$
    – John_dydx
    Jul 10, 2015 at 21:12
  • $\begingroup$ @John Thanks. And enjoy yours too. $\endgroup$
    – Mark Viola
    Jul 10, 2015 at 21:46
  • $\begingroup$ A bit behind on the limits 0 to 1-z $\endgroup$ Jul 11, 2015 at 15:27
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I have no idea of what you mean by type-1 or type-2, but in any case: $$\begin{eqnarray*}I=\int_{0}^{1}\int_{0}^{\log x}(x-1)\sqrt{1+e^{y}}\,dy\,dx &=& \int_{0}^{1}\int_{1}^{x}(x-1)\sqrt{1+z}\,\frac{dz}{z}\,dx\\&=&\int_{0}^{1}\int_{0}^{1-x}(1-x)\frac{\sqrt{2-z}}{1-z}\,dz\,dx\\&=&\int_{0}^{1}\int_{0}^{1-z}(1-x)\frac{\sqrt{2-z}}{1-z}\,dx\,dz\\&=&\int_{0}^{1}\frac{1-z^2}{2}\frac{\sqrt{2-z}}{1-z}\,dz\end{eqnarray*}$$ so: $$ I = \frac{1}{2}\int_{0}^{1}(1+z)\sqrt{2-z}\,dx = \frac{1}{2}\int_{1}^{2}(3-z)\sqrt{z}\,dx=\color{red}{\frac{2}{5}\left(3\sqrt{2}-2\right)}.$$

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  • $\begingroup$ More about type 1 and type 2 regions (Fubini's theorem)-users.math.msu.edu/users/gnagy/teaching/11-fall/mth234/…. $\endgroup$
    – John_dydx
    Jul 10, 2015 at 19:25
  • $\begingroup$ @John: oh, ok, so a type-1 region is a normal region with respect to the $x$-coordinate and a type-2 region is a normal region with respect to the $y$-coordinate. Since this problem boils down to an integration problem over the triangle $[0,1]^2\cap\{x+y\leq 1\}$, we have a region that is both type-1 and type-2. $\endgroup$ Jul 10, 2015 at 20:23
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    $\begingroup$ A little different from mine. Lots of ways to skin a cat. Enjoy the weekend. $\endgroup$
    – Mark Viola
    Jul 10, 2015 at 20:36
  • $\begingroup$ @JackD'Aurizio, precisely! $\endgroup$
    – John_dydx
    Jul 10, 2015 at 21:06
  • $\begingroup$ A bit behind on the limits 0 to 1-z $\endgroup$ Jul 11, 2015 at 15:28

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