What is the integral from 0 to y^3 of xe^{(-x)/y} ?

The integral from 0 to y^3 of xe^{(-x)/y} is -y^4e^{-y^2}-y^2e^{-y^2}+y^2

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integral from 0 to y^3 of xe^{(-x)/y}

\int_{0}^{y^{3}} x e^{\frac{- x}{y}} d x

- y^{4} e^{- y^{2}} - y^{2} e^{- y^{2}} + y^{2}

Solution steps

\int_{0}^{y^{3}} x e^{\frac{- x}{y}} d x

Simplify

= \int_{0}^{y^{3}} e^{- \frac{x}{y}} x d x

Apply Integration By Parts

= [- y e^{- \frac{x}{y}} x - \int - y e^{- \frac{x}{y}} d x]_{0}^{y^{3}}

\int - y e^{- \frac{x}{y}} d x = y^{2} e^{- \frac{x}{y}}

= [- y e^{- \frac{x}{y}} x - y^{2} e^{- \frac{x}{y}}]_{0}^{y^{3}}

Compute the boundaries:

- y^{4} e^{- y^{2}} - y^{2} e^{- y^{2}} + y^{2}

= - y^{4} e^{- y^{2}} - y^{2} e^{- y^{2}} + y^{2}

\int - 2 - \frac{2}{x} d x

\frac{d y}{d x} = \frac{2 x ^{2}}{y}

\frac{d}{d x} (2 x^{\frac{2}{3}})

d er i v a t i v e 2 + 3 x

\frac{d}{d x} (- 9 sin (3 x))

What is the integral from 0 to y^3 of xe^{(-x)/y} ?
The integral from 0 to y^3 of xe^{(-x)/y} is -y^4e^{-y^2}-y^2e^{-y^2}+y^2