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

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

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

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

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

Decimal

0.24999 \dots

Solution steps

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

Apply u-substitution

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 16}^{0} - \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_{- 16}^{0} e^{u} d u)

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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