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

The integral from-1 to 1 of (e^{2x}) is (e^4-1)/(2e^2)

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integral from-1 to 1 of (e^{2x})

\int_{- 1}^{1} (e^{2 x}) d x

\frac{e ^{4} - 1}{2 e ^{2}}

Decimal

3.62686 \dots

Solution steps

\int_{- 1}^{1} e^{2 x} d x

Apply u-substitution

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

= \int_{- 2}^{2} 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}^{2} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

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

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

Simplify

= \frac{e ^{4} - 1}{2 e ^{2}}

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What is the integral from-1 to 1 of (e^{2x}) ?
The integral from-1 to 1 of (e^{2x}) is (e^4-1)/(2e^2)