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

The integral from 1 to 2 of (e^{2x+1}) is (e^5-e^3)/2

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

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

\frac{e ^{5} - e ^{3}}{2}

Decimal

64.16381 \dots

Solution steps

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

Apply u-substitution

= \int_{3}^{5} 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_{3}^{5} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{3}^{5}

Compute the boundaries:

e^{5} - e^{3}

= \frac{1}{2} (e^{5} - e^{3})

Simplify

= \frac{e ^{5} - e ^{3}}{2}

What is the integral from 1 to 2 of (e^{2x+1}) ?
The integral from 1 to 2 of (e^{2x+1}) is (e^5-e^3)/2