What is the integral from 0 to 2 of (e^x)^2 ?

The integral from 0 to 2 of (e^x)^2 is (e^4-1)/2

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integral from 0 to 2 of (e^x)^2

\int_{0}^{2} (e^{x})^{2} d x

\frac{e ^{4} - 1}{2}

Decimal Notation

26.79907 \dots

Solution steps

\int_{0}^{2} (e^{x})^{2} d x

Apply u-substitution

: \int_{1}^{e^{2}} u d u

= \int_{1}^{e^{2}} u d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{e ^{4}}{2} - \frac{1}{2}

= \frac{e ^{4}}{2} - \frac{1}{2}

Simplify

= \frac{e ^{4} - 1}{2}

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What is the integral from 0 to 2 of (e^x)^2 ?
The integral from 0 to 2 of (e^x)^2 is (e^4-1)/2