What is the integral from 0 to 2 of x^3*e^{x^4} ?

The integral from 0 to 2 of x^3*e^{x^4} is (e^{16}-1)/4

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integral from 0 to 2 of x^3*e^{x^4}

\int_{0}^{2} x^{3} \cdot e^{x^{4}} d x

\frac{e ^{16} - 1}{4}

Solution steps

\int_{0}^{2} x^{3} e^{x^{4}} d x

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int_{0}^{8} \frac{e ^{u^{\frac{4}{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{4}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int_{0}^{16} \frac{3 e ^{v}}{4} d v

= \frac{1}{3} \cdot \int_{0}^{16} \frac{3 e ^{v}}{4} 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}{4} \cdot \int_{0}^{16} e^{v} d v

Use the common integral:

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

= \frac{1}{3} \cdot \frac{3}{4} [e^{v}]_{0}^{16}

Simplify

\frac{1}{3} \cdot \frac{3}{4} [e^{v}]_{0}^{16} : \frac{1}{4} [e^{v}]_{0}^{16}

= \frac{1}{4} [e^{v}]_{0}^{16}

Compute the boundaries:

e^{16} - 1

= \frac{1}{4} (e^{16} - 1)

Simplify

= \frac{e ^{16} - 1}{4}

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What is the integral from 0 to 2 of x^3*e^{x^4} ?
The integral from 0 to 2 of x^3*e^{x^4} is (e^{16}-1)/4