What is the integral from 1 to 4 of e^{2x+1} ?

The integral from 1 to 4 of e^{2x+1} is (e^9-e^3)/2

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integral from 1 to 4 of e^{2x+1}

\int_{1}^{4} e^{2 x + 1} d x

\frac{e ^{9} - e ^{3}}{2}

Decimal

4041.49919 \dots

Solution steps

\int_{1}^{4} e^{2 x + 1} d x

Apply u-substitution

: \int_{3}^{9} e^{u} \frac{1}{2} d u

= \int_{3}^{9} 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}^{9} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{3}^{9}

Compute the boundaries:

e^{9} - e^{3}

= \frac{1}{2} (e^{9} - e^{3})

Simplify

= \frac{e ^{9} - e ^{3}}{2}

\frac{\partial}{\partial y} (5 e^{x} cos (yz))

\int \frac{1}{x - 2 x x} d x

\int \frac{e ^{3 x}}{x ^{2}} d x

x \to 6 lim (\frac{1}{x ^{2} - 36})

(t - cos (t))^{^{'}}

What is the integral from 1 to 4 of e^{2x+1} ?
The integral from 1 to 4 of e^{2x+1} is (e^9-e^3)/2