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Can someone please show me the steps to solve $$ \frac{\sqrt{5}}{2}\int_{-1}^{1}xe^{-\frac{1}{2}x+\frac{3}{2}}dx $$ with integration by parts. I did it twice and got all different answers. I know the answer has to be -1.712. Thanks for your help

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It is unlikely that that is the answer. At most, that is an approximation to the answer. It is good practice not to approximate until you really need it. – Mlazhinka Shung Gronzalez LeWy Jul 6 '13 at 13:36…. that is the answer – jain smit Jul 6 '13 at 13:37
@jainsmit I believe Franklin's comment was a suggestion that you not approximate $\sqrt 5(e -3)e$ unless asked to. Note that in your image, the $\approx$ symbol means exactly this: that it is approximately equal to an infinitely long decimal expansion, for which $-1.712$ is even "less" of an approximation. Why not use the exact answer $\;\sqrt 5(e - 3)e$? – amWhy Jul 6 '13 at 13:48
All I am saying is to look carefully at the symbol $\approx$. It is more useful when it is a $=$. – Mlazhinka Shung Gronzalez LeWy Jul 6 '13 at 13:48
ok thanks for your help – jain smit Jul 6 '13 at 13:54


$$=\frac{\sqrt{5}}{2}\cdot e^{\frac32}\int_{-1}^1xe^{-\frac12x}dx$$

Putting $x=2y,dx=2dy$

When $x=-1,y=\frac x2=-\frac12$ and when $x=1,y=\frac x2=\frac12$

$$\implies I=\frac{\sqrt{5}}{2}e^{\frac32}\int_{-\frac12}^{\frac12}2ye^{-y}2dy=2\sqrt5e^{\frac32}\int_{-\frac12}^{\frac12}ye^{-y}dy$$

Now, $$ \begin{align} \int ye^{-y}dy &= y\left(\int e^{-y}dy\right)- \int\left(\frac{dy}{dy}\cdot\int e^{-y}dy\right)dy\\ &=-ye^{-y}+\int e^{-y}dy\\ &=-ye^{-y}-e^{-y} \end{align} $$

Can you take it from here?

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why did you make x = 2y? – jain smit Jul 6 '13 at 13:36
@jainsmit Presumably to get a pure exponential in the integrand. – Arthur Jul 6 '13 at 13:37
@jainsmit, as the power of $e$ is $-\frac x2$. We can set $y=-\frac x2$ as well – lab bhattacharjee Jul 6 '13 at 13:37
@labbhattacharjee im still not following you – jain smit Jul 6 '13 at 13:42
@jainsmit, have you noticed the power of $e$ in the integrand – lab bhattacharjee Jul 6 '13 at 13:46

$$ \int_{-1}^{1}xe^{-\frac{1}{2}x+\frac{3}{2}}dx =-2xe^{-\frac{1}{2}x+\frac{3}{2}}|_{x=-1}^{x=1} +2\int_{-1}^{1} e^{-\frac{1}{2}x+\frac{3}{2}}dx=-2e-2e^2-4 e^{-\frac{1}{2}x+\frac{3}{2}}|_{x=-1}^{x=1}$$

$$=-6e +2e^2.$$

We use the fact that $\int_a^bf'g = fg|_{a}^b-\int fg'$ and the fact that $\left(-2 e^{-\frac{1}{2}x+\frac{3}{2}}\right)'=e^{-\frac{1}{2}x+\frac{3}{2}}.$

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@Amzoti Multiplication by a constant is simple enough. The main problem was with integration by parts. I don't see why I should bloat the demonstration. – TZakrevskiy Jul 6 '13 at 13:42
you will with this poster as s/he is learning a lot of things and you must be explicit, but that was just a friendly comment, ignore if you feel not important. Regards – Amzoti Jul 6 '13 at 13:43

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