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Integrate e^(2x) on the interval [-1, 1]

Mathematics
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let u=2x so du=2dx or \(\large \frac{1}{2}du=dx \) so, \(\large \int_{-1}^{1}e^{2x}dx=\int_{-2}^{2}e^udu=e^u|_{-2}^{2}= \) ???
oops... there should be a factor of 1/2 outside of that second integral.....
So the limits do change..... was curious about that.... So it should be.... 27.2899172?

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this is what it should look like: \(\large \int_{-1}^{1}e^{2x}dx=\frac{1}{2}\int_{-2}^{2}e^udu=\frac{1}{2}e^u|_{-2}^{2}=\frac{1}{2}(e^2-e^{-2})=\frac{1}{2}(\frac{e^4-1}{e^2})=\frac{e^4-1}{2e^2} \) that's the exact form.
..... isn't u standing in for "2x".... therefore it would be e^4 - e^(-4)?
yes it is but we made a u-substitution that enables us to evaluate the integral in terms of u so we use -2 and 2 as the lower/upper limits of integration
if you wanted to back substitute so you could use x , then the limits would not change.
....Pretty sure in class, even when making u substitutions, we always would convert back to x form after integrating, keeping the new limits of the integral, and substituting those numbers in for x....
Fun fact: did you know you can actually perform definite integrals in scientific calculators?
Yes someone like a half a yr ago informed me. Just cant remember who
Uh, yeah.... but when I do it by hand I'm not getting the right answer. And I have to show my work as though I did it by hand.
Since this is a definite integral, the answer is numeric. so i don't see why you would back substitute back to x if we already made the transition from x to u. But either way, the numeric answer should be the same if you did a back subsitution to x or used the new limits involving u.

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