What is the integral from-1 to 2 of e^{3x} ?

The integral from-1 to 2 of e^{3x} is (e^9-1)/(3e^3)

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integral from-1 to 2 of e^{3x}

\int_{- 1}^{2} e^{3 x} d x

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

Decimal Notation

134.45966 \dots

Solution steps

\int_{- 1}^{2} e^{3 x} d x

Apply u-substitution

: \int_{- 3}^{6} e^{u} \frac{1}{3} d u

= \int_{- 3}^{6} 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}^{6} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{- 3}^{6}

Compute the boundaries:

e^{6} - \frac{1}{e ^{3}}

= \frac{1}{3} (e^{6} - \frac{1}{e ^{3}})

Simplify

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

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What is the integral from-1 to 2 of e^{3x} ?
The integral from-1 to 2 of e^{3x} is (e^9-1)/(3e^3)