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

The integral from 0 to infinity of y^2e^{-y} is 2

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integral from 0 to infinity of y^2e^{-y}

\int_{0}^{\infty} y^{2} e^{- y} d y

2

Solution steps

\int_{0}^{\infty} y^{2} e^{- y} d y

Compute the indefinite integral:

\int y^{2} e^{- y} d y = - e^{- y} y^{2} + 2 (- e^{- y} y - e^{- y}) + C

Compute the boundaries:

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

= 0 - (- 2)

Simplify

= 2

\int \frac{x ^{4}}{x ^{3} + x ^{2} - x - 1} d x

\int \frac{s}{2 s + 3} d s

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 2

\int x (3 + 5 x) d x

x \to 2 lim (3 f (x) + 2 g (x) (x)^{2})

What is the integral from 0 to infinity of y^2e^{-y} ?
The integral from 0 to infinity of y^2e^{-y} is 2