What is the integral from 0 to infinity of e^{-y}(2y^3-(2y^3)/3) ?

The integral from 0 to infinity of e^{-y}(2y^3-(2y^3)/3) is 8

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integral from 0 to infinity of e^{-y}(2y^3-(2y^3)/3)

\int_{0}^{\infty} e^{- y} (2 y^{3} - \frac{2 y ^{3}}{3}) d y

8

Solution steps

\int_{0}^{\infty} e^{- y} (2 y^{3} - \frac{2 y ^{3}}{3}) d y

Compute the indefinite integral:

\int e^{- y} (2 y^{3} - \frac{2 y ^{3}}{3}) d y = - \frac{4}{3} e^{- y} y^{3} - 4 e^{- y} y^{2} - 8 e^{- y} y - 8 e^{- y} + C

Compute the boundaries:

\int_{0}^{\infty} e^{- y} (2 y^{3} - \frac{2 y ^{3}}{3}) d y = 0 - (- 8)

= 0 - (- 8)

Simplify

= 8

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What is the integral from 0 to infinity of e^{-y}(2y^3-(2y^3)/3) ?
The integral from 0 to infinity of e^{-y}(2y^3-(2y^3)/3) is 8