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

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

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

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

2

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= 0 - (- 2)

Simplify

= 2

y^{^{'}} = 0.0016 y (2800 - y)

\int \frac{( x - 5 )}{x ^{2} - 18 x + 82} d x

f (x) = 1.5 x^{2} + 2 x^{1.5}

n = 1 \sum \infty \frac{1}{n ^{4} + n ^{2}}

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

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