What is the integral from 0 to 1 of e^{11x-10} ?

The integral from 0 to 1 of e^{11x-10} is (e^{11}-1)/(11e^{10)}

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integral from 0 to 1 of e^{11x-10}

\int_{0}^{1} e^{11 x - 10} d x

\frac{e ^{11} - 1}{11 e ^{10}}

Decimal

0.24711 \dots

Solution steps

\int_{0}^{1} e^{11 x - 10} d x

Apply u-substitution

: \int_{- 10}^{1} e^{u} \frac{1}{11} d u

= \int_{- 10}^{1} e^{u} \frac{1}{11} d u

Take the constant out:

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

= \frac{1}{11} \cdot \int_{- 10}^{1} e^{u} d u

Use the common integral:

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

= \frac{1}{11} [e^{u}]_{- 10}^{1}

Compute the boundaries:

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

= \frac{1}{11} (e - \frac{1}{e ^{10}})

Simplify

= \frac{e ^{11} - 1}{11 e ^{10}}

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What is the integral from 0 to 1 of e^{11x-10} ?
The integral from 0 to 1 of e^{11x-10} is (e^{11}-1)/(11e^{10)}