# Integration

#### matt07

##### New member
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

du

#### Dr.Peterson

##### Elite Member
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

##### Elite Member
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

##### Elite Member
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

##### Elite Member
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.