What is the integral from 0 to 1 of 4e^{5sqrt(x)} ?

The integral from 0 to 1 of 4e^{5sqrt(x)} is (8(4e^5+1))/(25)

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integral from 0 to 1 of 4e^{5sqrt(x)}

\int_{0}^{1} 4 e^{5 x} d x

\frac{8 ( 4 e ^{5} + 1 )}{25}

Decimal Notation

190.28884 \dots

Solution steps

\int_{0}^{1} 4 e^{5 x} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} e^{5 x} d x

Apply u-substitution

: \int_{0}^{5} \frac{2 e ^{u} u}{25} d u

= 4 \cdot \int_{0}^{5} \frac{2 e ^{u} u}{25} d u

Take the constant out:

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

= 4 \cdot \frac{2}{25} \cdot \int_{0}^{5} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{5}

= 4 \cdot \frac{2}{25} [e^{u} u - \int e^{u} d u]_{0}^{5}

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

= 4 \cdot \frac{2}{25} [e^{u} u - e^{u}]_{0}^{5}

Simplify

4 \cdot \frac{2}{25} [e^{u} u - e^{u}]_{0}^{5} : \frac{8}{25} [e^{u} u - e^{u}]_{0}^{5}

= \frac{8}{25} [e^{u} u - e^{u}]_{0}^{5}

Compute the boundaries:

4 e^{5} + 1

= \frac{8}{25} (4 e^{5} + 1)

Simplify

= \frac{8 ( 4 e ^{5} + 1 )}{25}

\int 2 - 4 x d x

\frac{d ^{2}}{d x ^{2}} (e^{9 e^{x}})

f (x) = \frac{5}{x + 5}

e^{\frac{x}{3}}

\int \frac{sin ( 3 x )}{cos ^{2} ( 3 x )} d x

What is the integral from 0 to 1 of 4e^{5sqrt(x)} ?
The integral from 0 to 1 of 4e^{5sqrt(x)} is (8(4e^5+1))/(25)