What is the integral from 0 to 1 of 6x^4e^{-x} ?

The integral from 0 to 1 of 6x^4e^{-x} is 6(-(65)/e+24)

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integral from 0 to 1 of 6x^4e^{-x}

\int_{0}^{1} 6 x^{4} e^{- x} d x

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

Decimal Notation

0.52701 \dots

Solution steps

\int_{0}^{1} 6 x^{4} e^{- x} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} x^{4} e^{- x} d x

Apply Integration By Parts

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

= 6 [- 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})))

= 6 [- e^{- x} x^{4} - (- 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x}))))]_{0}^{1}

Simplify

= 6 [- 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

= 6 (- \frac{65}{e} + 24)

x \to 0 lim (\frac{e ^{- 9 x} - 1}{x})

x \to 3 lim (5)

\int_{- 2}^{3} ∣ x + 1 ∣ d x

y^{^{'}} = \frac{5 x ^{2} + y ^{2}}{x y}

\int_{1}^{2} t^{2} t^{3} + 1 d t

What is the integral from 0 to 1 of 6x^4e^{-x} ?
The integral from 0 to 1 of 6x^4e^{-x} is 6(-(65)/e+24)