What is the integral from 0 to 1 of 5x^4e^{-x} ?

The integral from 0 to 1 of 5x^4e^{-x} is 5(-(65)/e+24)

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integral from 0 to 1 of 5x^4e^{-x}

\int_{0}^{1} 5 x^{4} e^{- x} d x

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

Solution steps

\int_{0}^{1} 5 x^{4} e^{- x} d x

Take the constant out:

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

= 5 \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}

= 5 [- 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})))

= 5 [- e^{- x} x^{4} - (- 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x}))))]_{0}^{1}

Simplify

= 5 [- 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

= 5 (- \frac{65}{e} + 24)

\int \frac{4}{x ^{2} - 14 x + 49} d x

\frac{\partial}{\partial x} (cos^{2} (5 x))

\frac{d}{d x} (x^{- \frac{3}{7}})

5 y^{^{''}} + y^{^{'}} = - 6 x, y (0) = 0, y^{^{'}} (0) = - 20

f (x) = 6 tan (x), at x = \frac{π}{3}

What is the integral from 0 to 1 of 5x^4e^{-x} ?
The integral from 0 to 1 of 5x^4e^{-x} is 5(-(65)/e+24)