- Thread starter matt07
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Have tried (or tried reviewing) substitution? Or parts? You may need both.

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

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 \(\displaystyle \int_{10}^{15}\frac{\pi}{2}e^{-\frac{\pi}{2}}du=\)\(\displaystyle \frac{\pi}{2}e^{-\frac{\pi}{2}}\int_{10}^{15} du=\)\(\displaystyle \frac{\pi}{2}e^{-\frac{\pi}{2}}(15- 10)\).

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

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.