What is the integral from 0 to infinity of 3ye^{-3y} ?

The integral from 0 to infinity of 3ye^{-3y} is 1/3

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integral from 0 to infinity of 3ye^{-3y}

\int_{0}^{\infty} 3 y e^{- 3 y} d y

\frac{1}{3}

Solution steps

\int_{0}^{\infty} 3 y e^{- 3 y} d y

Compute the indefinite integral:

\int 3 y e^{- 3 y} d y = - e^{- 3 y} y - \frac{1}{3} e^{- 3 y} + C

Compute the boundaries:

\int_{0}^{\infty} 3 y e^{- 3 y} d y = 0 - (- \frac{1}{3})

= 0 - (- \frac{1}{3})

Simplify

= \frac{1}{3}

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What is the integral from 0 to infinity of 3ye^{-3y} ?
The integral from 0 to infinity of 3ye^{-3y} is 1/3