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

The integral from 0 to 1 of (e^{2x+1})/2 is (e^3-e)/4

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integral from 0 to 1 of (e^{2x+1})/2

\int_{0}^{1} \frac{e ^{2 x + 1}}{2} d x

\frac{e ^{3} - e}{4}

Decimal Notation

4.34181 \dots

Solution steps

\int_{0}^{1} \frac{e ^{2 x + 1}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} e^{2 x + 1} d x

Apply u-substitution

: \int_{1}^{3} e^{u} \frac{1}{2} d u

= \frac{1}{2} \cdot \int_{1}^{3} 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 \frac{1}{2} \cdot \int_{1}^{3} e^{u} d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} [e^{u}]_{1}^{3}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [e^{u}]_{1}^{3} : \frac{1}{4} [e^{u}]_{1}^{3}

= \frac{1}{4} [e^{u}]_{1}^{3}

Compute the boundaries:

e^{3} - e

= \frac{1}{4} (e^{3} - e)

Simplify

= \frac{e ^{3} - e}{4}

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What is the integral from 0 to 1 of (e^{2x+1})/2 ?
The integral from 0 to 1 of (e^{2x+1})/2 is (e^3-e)/4