What is the integral from 0 to 2 of 2xe^{-x^2} ?

The integral from 0 to 2 of 2xe^{-x^2} is 1-1/(e^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 2xe^{-x^2}

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x e^{- x^{2}} d x

Apply u-substitution

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

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

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- 4}^{0})) : [e^{u}]_{- 4}^{0}

= [e^{u}]_{- 4}^{0}

Compute the boundaries:

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

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

\int_{- 2}^{2} (8 - 2 x^{2}) d x

\int_{0}^{3} \frac{x}{3} d x

\int_{- 1}^{3} \frac{x}{1 + ∣ x ∣} d x

\int_{0}^{2} k x d x

\int_{- 1}^{4} 3 x + 4 - x^{2} d x

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