What is the integral from 1 to 3 of e^{3x} ?

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

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integral from 1 to 3 of e^{3x}

\int_{1}^{3} e^{3 x} d x

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

Decimal Notation

2694.33279 \dots

Solution steps

\int_{1}^{3} e^{3 x} d x

Apply u-substitution

: \int_{3}^{9} e^{u} \frac{1}{3} d u

= \int_{3}^{9} e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{3}^{9} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{3}^{9}

Compute the boundaries:

e^{9} - e^{3}

= \frac{1}{3} (e^{9} - e^{3})

Simplify

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

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