What is the integral from 0 to 1 of e^{14x-13} ?

The integral from 0 to 1 of e^{14x-13} is (e^{14}-1)/(14e^{13)}

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integral from 0 to 1 of e^{14x-13}

\int_{0}^{1} e^{14 x - 13} d x

\frac{e ^{14} - 1}{14 e ^{13}}

Decimal

0.19416 \dots

Solution steps

\int_{0}^{1} e^{14 x - 13} d x

Apply u-substitution

: \int_{- 13}^{1} e^{u} \frac{1}{14} d u

= \int_{- 13}^{1} e^{u} \frac{1}{14} d u

Take the constant out:

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

= \frac{1}{14} \cdot \int_{- 13}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{14} [e^{u}]_{- 13}^{1}

Compute the boundaries:

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

= \frac{1}{14} (e - \frac{1}{e ^{13}})

Simplify

= \frac{e ^{14} - 1}{14 e ^{13}}

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What is the integral from 0 to 1 of e^{14x-13} ?
The integral from 0 to 1 of e^{14x-13} is (e^{14}-1)/(14e^{13)}