What is the integral from 2 to infinity of ye^{-4y} ?

The integral from 2 to infinity of ye^{-4y} is 9/(16e^8)

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integral from 2 to infinity of ye^{-4y}

\int_{2}^{\infty} y e^{- 4 y} d y

\frac{9}{16 e ^{8}}

Decimal Notation

0.00018 \dots

Solution steps

\int_{2}^{\infty} y e^{- 4 y} d y

Compute the indefinite integral:

\int y e^{- 4 y} d y = \frac{1}{16} (- 4 e^{- 4 y} y - e^{- 4 y}) + C

Compute the boundaries:

\int_{2}^{\infty} y e^{- 4 y} d y = 0 - (- \frac{9}{16 e ^{8}})

= 0 - (- \frac{9}{16 e ^{8}})

Simplify

= \frac{9}{16 e ^{8}}

\int arcsin (7 x) d x

\frac{d}{d x} (arctan (\frac{π}{x}))

\frac{1}{tan ( x )}

\frac{\partial}{\partial y} (5 + x ln (x y - 7))

x^{2} + \frac{1}{x ^{2}}

What is the integral from 2 to infinity of ye^{-4y} ?
The integral from 2 to infinity of ye^{-4y} is 9/(16e^8)