What is the integral from 0 to 1 of e^{13x-12} ?

The integral from 0 to 1 of e^{13x-12} is (e^{13}-1)/(13e^{12)}

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integral from 0 to 1 of e^{13x-12}

\int_{0}^{1} e^{13 x - 12} d x

\frac{e ^{13} - 1}{13 e ^{12}}

Decimal

0.20909 \dots

Solution steps

\int_{0}^{1} e^{13 x - 12} d x

Apply u-substitution

: \int_{- 12}^{1} e^{u} \frac{1}{13} d u

= \int_{- 12}^{1} e^{u} \frac{1}{13} d u

Take the constant out:

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

= \frac{1}{13} \cdot \int_{- 12}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{13} [e^{u}]_{- 12}^{1}

Compute the boundaries:

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

= \frac{1}{13} (e - \frac{1}{e ^{12}})

Simplify

= \frac{e ^{13} - 1}{13 e ^{12}}

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What is the integral from 0 to 1 of e^{13x-12} ?
The integral from 0 to 1 of e^{13x-12} is (e^{13}-1)/(13e^{12)}