integration by parts

May 2010
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0
Am i doing something wrong? I cant seem to spot my mistake, the answer is in red (given by my teacher).
 

skeeter

MHF Helper
Jun 2008
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North Texas
\(\displaystyle \int x^3 \cdot e^{\frac{x^2}{2}} \, dx\)

\(\displaystyle t = \frac{x^2}{2}\)

\(\displaystyle dt = x \, dx\)

\(\displaystyle 2 \int x \cdot \frac{x^2}{2} \cdot e^{\frac{x^2}{2}} \, dx\)

substitute ...

\(\displaystyle 2 \int t \cdot e^{t} \, dt\)

now do parts
 
May 2010
39
0
\(\displaystyle \int x^3 \cdot e^{\frac{x^2}{2}} \, dx\)

\(\displaystyle t = \frac{x^2}{2}\)

\(\displaystyle dt = x \, dx\)

\(\displaystyle 2 \int x \cdot \frac{x^2}{2} \cdot e^{\frac{x^2}{2}} \, dx\)

substitute ...

\(\displaystyle 2 \int t \cdot e^{t} \, dt\)

now do parts
thanks! just wondering, where did i go wrong in my first answer?
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
thanks! just wondering, where did i go wrong in my first answer?
\(\displaystyle e^{\frac{x^2}{2}}\) does not have an elementary antiderivative ... so, it cannot be chosen as "dv"
 
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HallsofIvy

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Apr 2005
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While, in differentiating something like \(\displaystyle e^{x^2}\) you can just multiply by the derivative of \(\displaystyle x^2\), to get \(\displaystyle 2xe^{x^2}\), when integrating you cannot just divide by the derivative- it must already be in the integral.
 
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May 2010
39
0
thanks everyone, very helpful! erm, could someone clarify for me: how do i know if something has an elementary antiderivative or not?
 
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