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

The integral from 0 to sqrt(2 of)x^3e^{-x^2} is -(-e^2+3)/(2e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{- e ^{2} + 3}{2 e ^{2}}

Decimal Notation

0.29699 \dots

Solution steps

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

Apply u-substitution

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

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

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

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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{- 2}^{0}

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

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

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

Compute the boundaries:

- 1 + \frac{3}{e ^{2}}

= - \frac{1}{2} (- 1 + \frac{3}{e ^{2}})

Simplify

= - \frac{- e ^{2} + 3}{2 e ^{2}}

\int_{0}^{2} \frac{y}{2} d y

\int_{1}^{4} (x^{2} - 1) d x

\int_{0}^{x} y cos (x^{2}) d y

\int_{0}^{\frac{1}{2}} (arcsin (x))^{2} d x

\int_{0}^{2 π} 4 cos^{2} (θ) d θ

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