What is the integral from 0 to 1 of x^3e^{x^2} ?

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

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integral from 0 to 1 of x^3e^{x^2}

\int_{0}^{1} x^{3} e^{x^{2}} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{0}^{1} x^{3} e^{x^{2}} d x

Apply u-substitution

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

Apply u-substitution

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

1

= \frac{1}{2} \cdot 1

Simplify

= \frac{1}{2}

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