What is the integral from 0 to 1 of 3x^4e^{-x} ?

The integral from 0 to 1 of 3x^4e^{-x} is 3(-(65)/e+24)

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integral from 0 to 1 of 3x^4e^{-x}

\int_{0}^{1} 3 x^{4} e^{- x} d x

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

Decimal

0.26350 \dots

Solution steps

\int_{0}^{1} 3 x^{4} e^{- x} d x

Take the constant out:

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

= 3 \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}

= 3 [- 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})))

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

Simplify

= 3 [- 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

= 3 (- \frac{65}{e} + 24)

y = - 5 x^{2}, (1, - 5)

20 ln (3 x + 1) - \frac{x}{3}

y = 8 x - x^{2}

\int \frac{1}{2 e ^{x}} d x

\int_{0}^{8} \frac{x}{x ^{2} + 36} d x

What is the integral from 0 to 1 of 3x^4e^{-x} ?
The integral from 0 to 1 of 3x^4e^{-x} is 3(-(65)/e+24)