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

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

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

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

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

Decimal

2221527.38012 \dots

Solution steps

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

Apply u-substitution

= \int_{0}^{16} \frac{e ^{u}}{4} d u

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{16} - 1

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

Simplify

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

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