What is the integral from 0 to 1 of 2x^4e^{-x} ?

The integral from 0 to 1 of 2x^4e^{-x} is 2(-(65)/e+24)

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integral from 0 to 1 of 2x^4e^{-x}

\int_{0}^{1} 2 x^{4} e^{- x} d x

2 (- \frac{65}{e} + 24)

Decimal

0.17567 \dots

Solution steps

\int_{0}^{1} 2 x^{4} 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}^{1} x^{4} e^{- x} d x

Apply Integration By Parts

= 2 [- e^{- x} x^{4} - \int - 4 e^{- x} x^{3} d x]_{0}^{1}

\int - 4 e^{- x} x^{3} d x = - 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x})))

= 2 [- e^{- x} x^{4} - (- 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x}))))]_{0}^{1}

Simplify

= 2 [- e^{- x} x^{4} + 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x})))]_{0}^{1}

Compute the boundaries:

- \frac{65}{e} + 24

= 2 (- \frac{65}{e} + 24)

s l o p e y = (x^{2} - 3)^{4}, at x = 2

\frac{d}{d x} (x^{\frac{x}{2}})

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What is the integral from 0 to 1 of 2x^4e^{-x} ?
The integral from 0 to 1 of 2x^4e^{-x} is 2(-(65)/e+24)