What is the integral from 2 to infinity of ye^{-8y} ?

The integral from 2 to infinity of ye^{-8y} is (17)/(64e^{16)}

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integral from 2 to infinity of ye^{-8y}

\int_{2}^{\infty} y e^{- 8 y} d y

\frac{17}{64 e ^{16}}

Decimal Notation

2.98922 E - 8

Solution steps

\int_{2}^{\infty} y e^{- 8 y} d y

Compute the indefinite integral:

\int y e^{- 8 y} d y = \frac{1}{64} (- 8 e^{- 8 y} y - e^{- 8 y}) + C

Compute the boundaries:

\int_{2}^{\infty} y e^{- 8 y} d y = 0 - (- \frac{17}{64 e ^{16}})

= 0 - (- \frac{17}{64 e ^{16}})

Simplify

= \frac{17}{64 e ^{16}}

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What is the integral from 2 to infinity of ye^{-8y} ?
The integral from 2 to infinity of ye^{-8y} is (17)/(64e^{16)}