What is the integral from 0 to 1 of (e^{3ax})/3 ?

The integral from 0 to 1 of (e^{3ax})/3 is (e^{3a}-1)/(9a)

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integral from 0 to 1 of (e^{3ax})/3

\int_{0}^{1} \frac{e ^{3 a x}}{3} d x

\frac{e ^{3 a} - 1}{9 a}

Solution steps

\int_{0}^{1} \frac{e ^{3 a x}}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} e^{3 a x} d x

Apply u-substitution

: \int_{0}^{3 a} \frac{e ^{u}}{3 a} d u

= \frac{1}{3} \cdot \int_{0}^{3 a} \frac{e ^{u}}{3 a} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{3 a} \cdot \int_{0}^{3 a} e^{u} d u

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{3 a} [e^{u}]_{0}^{3 a}

Simplify

\frac{1}{3} \cdot \frac{1}{3 a} [e^{u}]_{0}^{3 a} : \frac{[ e ^{u} ] _{0}^{3 a}}{9 a}

= \frac{[ e ^{u} ] _{0}^{3 a}}{9 a}

Compute the boundaries:

e^{3 a} - 1

= \frac{e ^{3 a} - 1}{9 a}

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What is the integral from 0 to 1 of (e^{3ax})/3 ?
The integral from 0 to 1 of (e^{3ax})/3 is (e^{3a}-1)/(9a)