What is the integral from 1 to 2 of 6e^{6x} ?

The integral from 1 to 2 of 6e^{6x} is e^{12}-e^6

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integral from 1 to 2 of 6e^{6x}

\int_{1}^{2} 6 e^{6 x} d x

e^{12} - e^{6}

Decimal

162351.36262 \dots

Solution steps

\int_{1}^{2} 6 e^{6 x} d x

Take the constant out:

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

= 6 \cdot \int_{1}^{2} e^{6 x} d x

Apply u-substitution

= 6 \cdot \int_{6}^{12} e^{u} \frac{1}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int_{6}^{12} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{6} [e^{u}]_{6}^{12}

Simplify

6 \cdot \frac{1}{6} [e^{u}]_{6}^{12} : [e^{u}]_{6}^{12}

= [e^{u}]_{6}^{12}

Compute the boundaries:

e^{12} - e^{6}

= e^{12} - e^{6}

What is the integral from 1 to 2 of 6e^{6x} ?
The integral from 1 to 2 of 6e^{6x} is e^{12}-e^6