Integration

matt07

New member
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May 22, 2020
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3
Hi,

Sorry for asking for help, im currently working through calculating energy but Im having difficulty integrating the below. I haven't touched integration for a long time and really struggling here. Is anyone able to help?

Thanks

1593474238099.pngdu
 

Dr.Peterson

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Nov 12, 2017
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8,036
That's a little hard to read, so I'm not sure which things are "u" and which are not.

Have tried (or tried reviewing) substitution? Or parts? You may need both.
 

Jomo

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Dec 30, 2014
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From what I see all the u's cancels out so in the end you are integrating a constant. Do you know how to integrate a constant? Either this integral is quite easy or we are not seeing the entire integral.
 

HallsofIvy

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Jan 27, 2012
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6,293
Echoing Dr. Peterson and Jomo, the problem looks like
\(\displaystyle \int_{10}^{15}\frac{\pi U}{2U}e^{-\pi\left(\frac{U}{2U}\right)^2}du\)

I not sure if what looks like "U", a capital U, is supposed to be "u", the variable in "du". But, in any case \(\displaystyle \frac{U}{2U}= \frac{1}{2}\) so the integral is just \(\displaystyle \int_{10}^{15}\frac{\pi}{2}e^{-\frac{\pi}{2}}du=\)\(\displaystyle \frac{\pi}{2}e^{-\frac{\pi}{2}}\int_{10}^{15} du=\)\(\displaystyle \frac{\pi}{2}e^{-\frac{\pi}{2}}(15- 10)\).
 

Dr.Peterson

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Nov 12, 2017
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I was wondering if it might really be something like this:

\(\displaystyle \int_{10}^{15}\frac{\pi u}{2U}e^{-\pi\left(\frac{u}{2U}\right)^2}du\)​

where U would be a constant. That's not very good form, but at least it would be doable without being trivial.

@matt07, why haven't you responded? If you had, we would have helped long ago.
 
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