What is the integral from 0 to 5 of 2x^2e^{-x} ?

The integral from 0 to 5 of 2x^2e^{-x} is 2(-(37)/(e^5)+2)

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integral from 0 to 5 of 2x^2e^{-x}

\int_{0}^{5} 2 x^{2} e^{- x} d x

2 (- \frac{37}{e ^{5}} + 2)

Decimal Notation

3.50139 \dots

Solution steps

\int_{0}^{5} 2 x^{2} e^{- x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{5} x^{2} e^{- x} d x

Apply Integration By Parts

: [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{0}^{5}

= 2 [- e^{- x} x^{2} - \int - 2 e^{- x} x d x]_{0}^{5}

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

= 2 [- e^{- x} x^{2} - (- 2 (- e^{- x} x - e^{- x}))]_{0}^{5}

Simplify

= 2 [- e^{- x} x^{2} + 2 (- e^{- x} x - e^{- x})]_{0}^{5}

Compute the boundaries:

- \frac{37}{e ^{5}} + 2

= 2 (- \frac{37}{e ^{5}} + 2)

3500 (1 - \frac{1}{50} t)^{2}

3 t (2 t^{2} - 5)^{4}

\frac{( s - 2 )}{( s - 2 ) ^{2} + 1}

x^{9} d x + y^{9} d y = 0

\frac{1}{s ( s ^{2} + 2 s + 1 )}

What is the integral from 0 to 5 of 2x^2e^{-x} ?
The integral from 0 to 5 of 2x^2e^{-x} is 2(-(37)/(e^5)+2)