What is the integral from 0 to 2 of 9e^{3x} ?

The integral from 0 to 2 of 9e^{3x} is 3(e^6-1)

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integral from 0 to 2 of 9e^{3x}

\int_{0}^{2} 9 e^{3 x} d x

3 (e^{6} - 1)

Solution steps

\int_{0}^{2} 9 e^{3 x} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{2} e^{3 x} d x

Apply u-substitution

: \int_{0}^{6} e^{u} \frac{1}{3} d u

= 9 \cdot \int_{0}^{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

= 9 \cdot \frac{1}{3} \cdot \int_{0}^{6} e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} [e^{u}]_{0}^{6}

Simplify

9 \cdot \frac{1}{3} [e^{u}]_{0}^{6} : 3 [e^{u}]_{0}^{6}

= 3 [e^{u}]_{0}^{6}

Compute the boundaries:

e^{6} - 1

= 3 (e^{6} - 1)

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What is the integral from 0 to 2 of 9e^{3x} ?
The integral from 0 to 2 of 9e^{3x} is 3(e^6-1)