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

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

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

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

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

Solution steps

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

Simplify

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

\int_{0}^{\frac{13}{6}} π (\frac{13}{6} - x)^{2} d x

\frac{\partial}{\partial x} (sin (π (x - 6 y)))

\int π (4 x)^{2} d x

\int \frac{x ^{2}}{9 + x ^{2}} d x

\frac{d}{d x} ((x^{2} - 4)^{\frac{1}{3}})

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