What is the integral from-e to 0 of e^{2x} ?

The integral from-e to 0 of e^{2x} is (e^{2e}-1)/(2e^{2e)}

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integral from-e to 0 of e^{2x}

\int_{- e}^{0} e^{2 x} d x

\frac{e ^{2 e} - 1}{2 e ^{2 e}}

Decimal Notation

0.49782 \dots

Solution steps

\int_{- e}^{0} e^{2 x} d x

Apply u-substitution

: \int_{- 2 e}^{0} e^{u} \frac{1}{2} d u

= \int_{- 2 e}^{0} e^{u} \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{- 2 e}^{0} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{- 2 e}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{2 e}}

= \frac{1}{2} (1 - \frac{1}{e ^{2 e}})

Simplify

= \frac{e ^{2 e} - 1}{2 e ^{2 e}}

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What is the integral from-e to 0 of e^{2x} ?
The integral from-e to 0 of e^{2x} is (e^{2e}-1)/(2e^{2e)}