What is the integral from 0 to 1 of e^{6x-5} ?

The integral from 0 to 1 of e^{6x-5} is (e^6-1)/(6e^5)

integral from 0 to 1 of e^{6x-5}

\int_{0}^{1} e^{6 x - 5} d x

\frac{e ^{6} - 1}{6 e ^{5}}

Decimal

0.45192 \dots

Solution steps

\int_{0}^{1} e^{6 x - 5} d x

Apply u-substitution

= \int_{- 5}^{1} 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

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

Use the common integral:

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

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

Compute the boundaries:

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

= \frac{1}{6} (e - \frac{1}{e ^{5}})

Simplify

= \frac{e ^{6} - 1}{6 e ^{5}}

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What is the integral from 0 to 1 of e^{6x-5} ?
The integral from 0 to 1 of e^{6x-5} is (e^6-1)/(6e^5)