What is the integral from 0 to 1 of 7x^4e^{-x} ?

The integral from 0 to 1 of 7x^4e^{-x} is 7(-(65)/e+24)

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integral from 0 to 1 of 7x^4e^{-x}

\int_{0}^{1} 7 x^{4} e^{- x} d x

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

Decimal Notation

0.61485 \dots

Solution steps

\int_{0}^{1} 7 x^{4} e^{- x} d x

Take the constant out:

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

= 7 \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}

= 7 [- 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})))

= 7 [- e^{- x} x^{4} - (- 4 (- e^{- x} x^{3} - 3 (e^{- x} x^{2} - 2 (- e^{- x} x - e^{- x}))))]_{0}^{1}

Simplify

= 7 [- 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

= 7 (- \frac{65}{e} + 24)

\int \frac{21}{21 + e ^{x}} d x

y^{^{'}} = \frac{3 x ^{2} + 2 x + 1}{y - 2}

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

\int \frac{9}{x ^{2}} d x

\frac{1}{10 s}

What is the integral from 0 to 1 of 7x^4e^{-x} ?
The integral from 0 to 1 of 7x^4e^{-x} is 7(-(65)/e+24)