What is the integral from 0 to 1 of 2e^{6x-1} ?

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

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integral from 0 to 1 of 2e^{6x-1}

\int_{0}^{1} 2 e^{6 x - 1} d x

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

Decimal Notation

49.34842 \dots

Solution steps

\int_{0}^{1} 2 e^{6 x - 1} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{1} e^{6 x - 1} d x

Apply u-substitution

: \int_{- 1}^{5} e^{u} \frac{1}{6} d u

= 2 \cdot \int_{- 1}^{5} 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

= 2 \cdot \frac{1}{6} \cdot \int_{- 1}^{5} e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{6} [e^{u}]_{- 1}^{5}

Simplify

2 \cdot \frac{1}{6} [e^{u}]_{- 1}^{5} : \frac{1}{3} [e^{u}]_{- 1}^{5}

= \frac{1}{3} [e^{u}]_{- 1}^{5}

Compute the boundaries:

e^{5} - \frac{1}{e}

= \frac{1}{3} (e^{5} - \frac{1}{e})

Simplify

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

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