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

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

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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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