What is the integral from 0 to 1 of x^4e^{3x^5} ?

The integral from 0 to 1 of x^4e^{3x^5} is (e^3-1)/(15)

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integral from 0 to 1 of x^4e^{3x^5}

\int_{0}^{1} x^{4} e^{3 x^{5}} d x

\frac{e ^{3} - 1}{15}

Decimal Notation

1.27236 \dots

Solution steps

\int_{0}^{1} x^{4} e^{3 x^{5}} d x

Apply u-substitution

: \int_{0}^{1} \frac{e ^{3 u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

= \int_{0}^{1} \frac{e ^{3 u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{1} e^{3 u^{\frac{5}{4}}} u^{\frac{1}{4}} d u

Apply u-substitution

: \int_{0}^{1} \frac{4 e ^{3 v}}{5} d v

= \frac{1}{4} \cdot \int_{0}^{1} \frac{4 e ^{3 v}}{5} d v

Take the constant out:

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

= \frac{1}{4} \cdot \frac{4}{5} \cdot \int_{0}^{1} e^{3 v} d v

Apply u-substitution

: \int_{0}^{3} e^{w} \frac{1}{3} d w

= \frac{1}{4} \cdot \frac{4}{5} \cdot \int_{0}^{3} e^{w} \frac{1}{3} d w

Take the constant out:

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

= \frac{1}{4} \cdot \frac{4}{5} \cdot \frac{1}{3} \cdot \int_{0}^{3} e^{w} d w

Use the common integral:

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

= \frac{1}{4} \cdot \frac{4}{5} \cdot \frac{1}{3} [e^{w}]_{0}^{3}

Simplify

\frac{1}{4} \cdot \frac{4}{5} \cdot \frac{1}{3} [e^{w}]_{0}^{3} : \frac{1}{15} [e^{w}]_{0}^{3}

= \frac{1}{15} [e^{w}]_{0}^{3}

Compute the boundaries:

e^{3} - 1

= \frac{1}{15} (e^{3} - 1)

Simplify

= \frac{e ^{3} - 1}{15}

\frac{d}{d x} (\frac{4}{5 + cos ( x )})

f (x) = 8 (x - \frac{1}{x})^{2}, at x = 2

\frac{d}{d x} (1 0^{- 3 x^{2}})

\int x^{2} + 49 d x

y = 2 x, y = \frac{2 x}{3}

What is the integral from 0 to 1 of x^4e^{3x^5} ?
The integral from 0 to 1 of x^4e^{3x^5} is (e^3-1)/(15)