What is the integral from 1 to 2 of xe^{-x^2} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of xe^{-x^2}

\int_{1}^{2} x e^{- x^{2}} d x

\frac{e ^{3} - 1}{2 e ^{4}}

Decimal Notation

0.17478 \dots

Solution steps

\int_{1}^{2} x e^{- x^{2}} d x

Apply u-substitution

: \int_{- 1}^{- 4} - \frac{e ^{u}}{2} d u

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

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

= - \int_{- 4}^{- 1} - \frac{e ^{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}^{- 1} e^{u} d u)

Use the common integral:

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

= \frac{e ^{3} - 1}{2 e ^{4}}

(0, 0), (- 5, 3)

\int csc^{2} (\frac{x}{2}) d x

\frac{s + 5}{s ^{2} + 4 s + 3}

\frac{d}{d x} (e^{x} + c)

f (x) = x^{2} + x + 1

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