What is the integral from 0 to 1 of e^{ax} ?

The integral from 0 to 1 of e^{ax} is (e^a-1)/a

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integral from 0 to 1 of e^{ax}

\int_{0}^{1} e^{a x} d x

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

Solution steps

\int_{0}^{1} e^{a x} d x

Apply u-substitution

: \int_{0}^{a} \frac{e ^{u}}{a} d u

= \int_{0}^{a} \frac{e ^{u}}{a} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a} \cdot \int_{0}^{a} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{a} [e^{u}]_{0}^{a}

Simplify

\frac{1}{a} [e^{u}]_{0}^{a} : \frac{[ e ^{u} ] _{0}^{a}}{a}

= \frac{[ e ^{u} ] _{0}^{a}}{a}

Compute the boundaries:

e^{a} - 1

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

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What is the integral from 0 to 1 of e^{ax} ?
The integral from 0 to 1 of e^{ax} is (e^a-1)/a