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

The integral from 0 to 2 of x^3e^{x^2} is (3e^4+1)/2

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

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

\frac{3 e ^{4} + 1}{2}

Decimal Notation

82.39722 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u^{\frac{2}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int_{0}^{8} \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}^{8} e^{u^{\frac{2}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int_{0}^{4} \frac{3 e ^{v} v}{2} d v

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

Apply Integration By Parts

: [e^{v} v - \int e^{v} d v]_{0}^{4}

= \frac{1}{3} \cdot \frac{3}{2} [e^{v} v - \int e^{v} d v]_{0}^{4}

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

= \frac{1}{3} \cdot \frac{3}{2} [e^{v} v - e^{v}]_{0}^{4}

Simplify

\frac{1}{3} \cdot \frac{3}{2} [e^{v} v - e^{v}]_{0}^{4} : \frac{1}{2} [e^{v} v - e^{v}]_{0}^{4}

= \frac{1}{2} [e^{v} v - e^{v}]_{0}^{4}

Compute the boundaries:

3 e^{4} + 1

= \frac{1}{2} (3 e^{4} + 1)

Simplify

= \frac{3 e ^{4} + 1}{2}

\int_{- 1}^{3} ∣ x - 2 ∣ d x

\frac{\partial}{\partial y} (2 x^{4} y z^{3})

cos (2 x)

\frac{d}{d x} (\frac{x y}{x ^{2} + y ^{2} + z ^{2}})

f (x) = ln ((2 x - 5) (3 x^{2} - 5))

What is the integral from 0 to 2 of x^3e^{x^2} ?
The integral from 0 to 2 of x^3e^{x^2} is (3e^4+1)/2