What is the integral from 0 to 5 of xe^{-x/4} ?

The integral from 0 to 5 of xe^{-x/4} is -(36)/(e^{5/4)}+16

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integral from 0 to 5 of xe^{-x/4}

\int_{0}^{5} x e^{- \frac{x}{4}} d x

- \frac{36}{e ^{\frac{5}{4}}} + 16

Decimal Notation

5.68582 \dots

Solution steps

\int_{0}^{5} x e^{- \frac{x}{4}} d x

Apply Integration By Parts

: [- 4 e^{- \frac{x}{4}} x - \int - 4 e^{- \frac{x}{4}} d x]_{0}^{5}

= [- 4 e^{- \frac{x}{4}} x - \int - 4 e^{- \frac{x}{4}} d x]_{0}^{5}

\int - 4 e^{- \frac{x}{4}} d x = 16 e^{- \frac{x}{4}}

= [- 4 e^{- \frac{x}{4}} x - 16 e^{- \frac{x}{4}}]_{0}^{5}

Compute the boundaries:

- \frac{36}{e ^{\frac{5}{4}}} + 16

= - \frac{36}{e ^{\frac{5}{4}}} + 16

\int_{0}^{4} x + 1 d x

\int_{1}^{\infty} \frac{1}{4 x - 1} d x

\int_{\frac{π}{2}}^{0} - sin^{2} (t) d t

\int_{0}^{1} x^{2} 7 x + 8 d x

\int_{0}^{1} x^{2} (\frac{1}{2} + \frac{3}{2} x^{2}) d x

What is the integral from 0 to 5 of xe^{-x/4} ?
The integral from 0 to 5 of xe^{-x/4} is -(36)/(e^{5/4)}+16