What is the integral from 0 to 4 of xe^{-ax} ?

The integral from 0 to 4 of xe^{-ax} is (-4ae^{-4a}-e^{-4a}+1)/(a^2)

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integral from 0 to 4 of xe^{-ax}

\int_{0}^{4} x e^{- a x} d x

\frac{- 4 a e ^{- 4 a} - e ^{- 4 a} + 1}{a ^{2}}

Solution steps

\int_{0}^{4} x e^{- a x} d x

Apply u-substitution

: \int_{0}^{- a \cdot 4} \frac{e ^{u} u}{a ^{2}} d u

= \int_{0}^{- a \cdot 4} \frac{e ^{u} u}{a ^{2}} d u

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int_{0}^{- a \cdot 4} e^{u} u d u

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- 4 a}

= \frac{1}{a ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- 4 a}

\int e^{u} d u = e^{u}

= \frac{1}{a ^{2}} [e^{u} u - e^{u}]_{0}^{- 4 a}

Simplify

\frac{1}{a ^{2}} [e^{u} u - e^{u}]_{0}^{- 4 a} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- 4 a}}{a ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- 4 a}}{a ^{2}}

Compute the boundaries:

- 4 a e^{- 4 a} - e^{- 4 a} + 1

= \frac{- 4 a e ^{- 4 a} - e ^{- 4 a} + 1}{a ^{2}}

e^{8 x}

\int x^{2} 81 - x^{2} d x

f (x) = x 9 - x

\int (3 x^{3} + 2 3 x^{2}) d x

\frac{1}{( s ^{2} + 1 ) ( s ^{3} + 1 )}

What is the integral from 0 to 4 of xe^{-ax} ?
The integral from 0 to 4 of xe^{-ax} is (-4ae^{-4a}-e^{-4a}+1)/(a^2)