What is the integral from-1 to 1 of (e^x)^2 ?

The integral from-1 to 1 of (e^x)^2 is (e^2)/2-1/(2e^2)

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integral from-1 to 1 of (e^x)^2

\int_{- 1}^{1} (e^{x})^{2} d x

\frac{e ^{2}}{2} - \frac{1}{2 e ^{2}}

Decimal

3.62686 \dots

Solution steps

\int_{- 1}^{1} (e^{x})^{2} d x

Apply u-substitution

: \int_{\frac{1}{e}}^{e} u d u

= \int_{\frac{1}{e}}^{e} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{\frac{1}{e}}^{e}

= [\frac{u ^{2}}{2}]_{\frac{1}{e}}^{e}

Compute the boundaries:

\frac{e ^{2}}{2} - \frac{1}{2 e ^{2}}

= \frac{e ^{2}}{2} - \frac{1}{2 e ^{2}}

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\frac{d}{d x} (\frac{2}{x y})

What is the integral from-1 to 1 of (e^x)^2 ?
The integral from-1 to 1 of (e^x)^2 is (e^2)/2-1/(2e^2)