Question 514162
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V\ =\ \int_{-1}^{\ \ \ \ \ \ 1}\ e^{\small{2x}}\LARGE dx\ =\ \left|\frac{e^{2x}}{2}\right|_{-1}^1\ =\ \frac{e^2}{2}\ -\ \frac{1}{2e^2}]


Numerical approximation 3.627


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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