Integration
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Dr.Peterson
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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.
Have tried (or tried reviewing) substitution? Or parts? You may need both.
HallsofIvy
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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 \(\displaystyle \int_{10}^{15}\frac{\pi}{2}e^{-\frac{\pi}{2}}du=\)\(\displaystyle \frac{\pi}{2}e^{-\frac{\pi}{2}}\int_{10}^{15} du=\)\(\displaystyle \frac{\pi}{2}e^{-\frac{\pi}{2}}(15- 10)\).
\(\displaystyle \int_{10}^{15}\frac{\pi U}{2U}e^{-\pi\left(\frac{U}{2U}\right)^2}du\)
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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I was wondering if it might really be something like this:
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.
[MATH]\int_{10}^{15}\frac{\pi u}{2U}e^{-\pi\left(\frac{u}{2U}\right)^2}du[/MATH]
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.