What is the integral from 0 to 1 of e^{5x-4} ?

The integral from 0 to 1 of e^{5x-4} is (e^5-1)/(5e^4)

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integral from 0 to 1 of e^{5x-4}

\int_{0}^{1} e^{5 x - 4} d x

\frac{e ^{5} - 1}{5 e ^{4}}

Solution steps

\int_{0}^{1} e^{5 x - 4} d x

Apply u-substitution

: \int_{- 4}^{1} e^{u} \frac{1}{5} d u

= \int_{- 4}^{1} e^{u} \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{- 4}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{5} [e^{u}]_{- 4}^{1}

Compute the boundaries:

e - \frac{1}{e ^{4}}

= \frac{1}{5} (e - \frac{1}{e ^{4}})

Simplify

= \frac{e ^{5} - 1}{5 e ^{4}}

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What is the integral from 0 to 1 of e^{5x-4} ?
The integral from 0 to 1 of e^{5x-4} is (e^5-1)/(5e^4)