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

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}

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\int 12 x^{2} sin (6 - x^{3}) d x

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