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

The integral from 0 to 3 of x^5e^{x^3} is (26e^{27}+1)/3

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

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

\frac{26 e ^{27} + 1}{3}

Decimal

4611084751882.581

Solution steps

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

Apply u-substitution

: \int_{0}^{243} \frac{e ^{u^{\frac{3}{5}}} u ^{\frac{1}{5}}}{5} d u

= \int_{0}^{243} \frac{e ^{u^{\frac{3}{5}}} u ^{\frac{1}{5}}}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{0}^{243} e^{u^{\frac{3}{5}}} u^{\frac{1}{5}} d u

Apply u-substitution

: \int_{0}^{27} \frac{5 e ^{v} v}{3} d v

= \frac{1}{5} \cdot \int_{0}^{27} \frac{5 e ^{v} v}{3} d v

Take the constant out:

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

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int_{0}^{27} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{27}

= \frac{1}{5} \cdot \frac{5}{3} [e^{v} v - \int e^{v} d v]_{0}^{27}

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

= \frac{1}{5} \cdot \frac{5}{3} [e^{v} v - e^{v}]_{0}^{27}

Simplify

\frac{1}{5} \cdot \frac{5}{3} [e^{v} v - e^{v}]_{0}^{27} : \frac{1}{3} [e^{v} v - e^{v}]_{0}^{27}

= \frac{1}{3} [e^{v} v - e^{v}]_{0}^{27}

Compute the boundaries:

26 e^{27} + 1

= \frac{1}{3} (26 e^{27} + 1)

Simplify

= \frac{26 e ^{27} + 1}{3}

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What is the integral from 0 to 3 of x^5e^{x^3} ?
The integral from 0 to 3 of x^5e^{x^3} is (26e^{27}+1)/3