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 -(-e^4+5)/(2e^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{- e ^{4} + 5}{2 e ^{4}}

Decimal

0.45421 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{- 4} \frac{e ^{u} u}{2} d u

= \int_{0}^{- 4} \frac{e ^{u} u}{2} d u

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

= - \int_{- 4}^{0} \frac{e ^{u} u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int_{- 4}^{0} e^{u} u d u

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

- 1 + \frac{5}{e ^{4}}

= - \frac{1}{2} (- 1 + \frac{5}{e ^{4}})

Simplify

= - \frac{- e ^{4} + 5}{2 e ^{4}}

x \to 0 lim (\frac{1 + x}{x ^{2}})

4 (x - (\frac{1}{x})^{4})

f (x) = \frac{x}{3}

\frac{d}{d x} (arccos (cos^{2} (x)))

\frac{d}{d x} (12 sin (x))

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 -(-e^4+5)/(2e^4)