What is the integral from 0 to 1 of e^{10x-9} ?

The integral from 0 to 1 of e^{10x-9} is (e^{10}-1)/(10e^9)

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integral from 0 to 1 of e^{10x-9}

\int_{0}^{1} e^{10 x - 9} d x

\frac{e ^{10} - 1}{10 e ^{9}}

Decimal

0.27181 \dots

Solution steps

\int_{0}^{1} e^{10 x - 9} d x

Apply u-substitution

: \int_{- 9}^{1} e^{u} \frac{1}{10} d u

= \int_{- 9}^{1} e^{u} \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int_{- 9}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{10} [e^{u}]_{- 9}^{1}

Compute the boundaries:

e - \frac{1}{e ^{9}}

= \frac{1}{10} (e - \frac{1}{e ^{9}})

Simplify

= \frac{e ^{10} - 1}{10 e ^{9}}

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What is the integral from 0 to 1 of e^{10x-9} ?
The integral from 0 to 1 of e^{10x-9} is (e^{10}-1)/(10e^9)