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

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

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

\int_{0}^{l n (2)} e^{2 x + 1} d x

\frac{3 e}{2}

Decimal

4.07742 \dots

Solution steps

\int_{0}^{l n (2)} e^{2 x + 1} d x

Apply u-substitution

: \int_{1}^{2 l n (2) + 1} e^{u} \frac{1}{2} d u

= \int_{1}^{2 l n (2) + 1} 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_{1}^{2 l n (2) + 1} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{1}^{2 l n (2) + 1}

Compute the boundaries:

3 e

= \frac{1}{2} \cdot 3 e

Simplify

= \frac{3 e}{2}

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