What is the integral from 0 to 1 of sqrt(e^{3x+1)} ?

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

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integral from 0 to 1 of sqrt(e^{3x+1)}

\int_{0}^{1} e^{3 x + 1} d x

\frac{2 ( e ^{2} - e )}{3}

Decimal Notation

3.82688 \dots

Solution steps

\int_{0}^{1} e^{3 x + 1} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{3} e^{\frac{u}{2}} d u

= \int_{1}^{4} \frac{1}{3} e^{\frac{u}{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{1}^{4} e^{\frac{u}{2}} d u

Apply u-substitution

: \int_{\frac{1}{2}}^{2} e^{v} \cdot 2 d v

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e^{2} - e

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

Simplify

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

\frac{d}{d x} (\frac{( x ^{2} + 4 )}{x})

y = 2 x^{3} - 8 x, (2, 0)

\frac{d v}{d t} = - v

\int s h \frac{x}{3} d x

f (x) = 4 x^{2} - 8, at 3

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