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

The integral from 5 to infinity of xe^{-x/2} is (14)/(e^{5/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{5}^{\infty} x e^{- \frac{x}{2}} d x

\frac{14}{e ^{\frac{5}{2}}}

Decimal Notation

1.14918 \dots

Solution steps

\int_{5}^{\infty} x e^{- \frac{x}{2}} d x

Compute the indefinite integral:

\int x e^{- \frac{x}{2}} d x = - 2 e^{- \frac{x}{2}} x - 4 e^{- \frac{x}{2}} + C

Compute the boundaries:

\int_{5}^{\infty} x e^{- \frac{x}{2}} d x = 0 - (- \frac{14}{e ^{\frac{5}{2}}})

= 0 - (- \frac{14}{e ^{\frac{5}{2}}})

Simplify

= \frac{14}{e ^{\frac{5}{2}}}

\int_{1}^{2} x^{3} - 2 x^{2} - 5 x + 6 d x

\int_{0}^{1} 2 y^{2} - 2 y d y

\int_{- 8}^{3} 1 + ∣ x ∣ d x

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

\int_{0}^{2 π} \frac{1}{1 + sin ^{2} ( x )} d x

What is the integral from 5 to infinity of xe^{-x/2} ?
The integral from 5 to infinity of xe^{-x/2} is (14)/(e^{5/2)}