What is the integral from-1 to 5 of 5xe^{-x^2} ?

The integral from-1 to 5 of 5xe^{-x^2} is (5(1/e-1/(e^{25)}))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from-1 to 5 of 5xe^{-x^2}

\int_{- 1}^{5} 5 x e^{- x^{2}} d x

\frac{5 ( \frac{1}{e} - \frac{1}{e ^{25}} )}{2}

Decimal

0.91969 \dots

Solution steps

\int_{- 1}^{5} 5 x e^{- x^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{- 1}^{5} x e^{- x^{2}} d x

Apply u-substitution

= 5 \cdot \int_{- 1}^{- 25} - \frac{e ^{u}}{2} d u

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

= 5 (- \int_{- 25}^{- 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

= 5 (- (- \frac{1}{2} \cdot \int_{- 25}^{- 1} e^{u} d u))

Use the common integral:

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

= 5 (- (- \frac{1}{2} [e^{u}]_{- 25}^{- 1}))

Simplify

5 (- (- \frac{1}{2} [e^{u}]_{- 25}^{- 1})) : \frac{5}{2} [e^{u}]_{- 25}^{- 1}

= \frac{5}{2} [e^{u}]_{- 25}^{- 1}

Compute the boundaries:

\frac{1}{e} - \frac{1}{e ^{25}}

= \frac{5}{2} (\frac{1}{e} - \frac{1}{e ^{25}})

Simplify

= \frac{5 ( \frac{1}{e} - \frac{1}{e ^{25}} )}{2}

What is the integral from-1 to 5 of 5xe^{-x^2} ?
The integral from-1 to 5 of 5xe^{-x^2} is (5(1/e-1/(e^{25)}))/2