What is the integral from 0 to 6 of e^{2x} ?

The integral from 0 to 6 of e^{2x} is (e^{12}-1)/2

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integral from 0 to 6 of e^{2x}

\int_{0}^{6} e^{2 x} d x

\frac{e ^{12} - 1}{2}

Decimal

81376.89570 \dots

Solution steps

\int_{0}^{6} e^{2 x} d x

Apply u-substitution

: \int_{0}^{12} e^{u} \frac{1}{2} d u

= \int_{0}^{12} e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{12} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{0}^{12}

Compute the boundaries:

e^{12} - 1

= \frac{1}{2} (e^{12} - 1)

Simplify

= \frac{e ^{12} - 1}{2}

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What is the integral from 0 to 6 of e^{2x} ?
The integral from 0 to 6 of e^{2x} is (e^{12}-1)/2