What is the integral from 0 to 3 of x^3e^{x^2} ?

The integral from 0 to 3 of x^3e^{x^2} is (8e^9+1)/2

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integral from 0 to 3 of x^3e^{x^2}

\int_{0}^{3} x^{3} e^{x^{2}} d x

\frac{8 e ^{9} + 1}{2}

Decimal Notation

32412.83571 \dots

Solution steps

\int_{0}^{3} x^{3} e^{x^{2}} d x

Apply u-substitution

: \int_{0}^{27} \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int_{0}^{27} \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} 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_{0}^{27} e^{u^{\frac{2}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int_{0}^{9} \frac{3 e ^{v} v}{2} d v

= \frac{1}{3} \cdot \int_{0}^{9} \frac{3 e ^{v} v}{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 \frac{3}{2} \cdot \int_{0}^{9} e^{v} v d v

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{9}

= \frac{1}{3} \cdot \frac{3}{2} [e^{v} v - \int e^{v} d v]_{0}^{9}

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

= \frac{1}{3} \cdot \frac{3}{2} [e^{v} v - e^{v}]_{0}^{9}

Simplify

\frac{1}{3} \cdot \frac{3}{2} [e^{v} v - e^{v}]_{0}^{9} : \frac{1}{2} [e^{v} v - e^{v}]_{0}^{9}

= \frac{1}{2} [e^{v} v - e^{v}]_{0}^{9}

Compute the boundaries:

8 e^{9} + 1

= \frac{1}{2} (8 e^{9} + 1)

Simplify

= \frac{8 e ^{9} + 1}{2}

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What is the integral from 0 to 3 of x^3e^{x^2} ?
The integral from 0 to 3 of x^3e^{x^2} is (8e^9+1)/2