What is the integral of xe^{4-x} ?

The integral of xe^{4-x} is -e^{4-x}x-e^{4-x}+C

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integral of xe^{4-x}

\int x e^{4 - x} d x

- e^{4 - x} x - e^{4 - x} + C

Solution steps

\int x e^{4 - x} d x

Apply u-substitution

: \int - e^{u} (- u + 4) d u

= \int - e^{u} (- u + 4) d u

Take the constant out:

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

= - \int e^{u} (- u + 4) d u

Apply Integration By Parts

: e^{u} (- u + 4) - \int - e^{u} d u

= - (e^{u} (- u + 4) - \int - e^{u} d u)

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

= - (e^{u} (- u + 4) - (- e^{u}))

Substitute back

u = 4 - x

= - (e^{4 - x} (- (4 - x) + 4) - (- e^{4 - x}))

Simplify

- (e^{4 - x} (- (4 - x) + 4) - (- e^{4 - x})) : - e^{4 - x} x - e^{4 - x}

= - e^{4 - x} x - e^{4 - x}

Add a constant to the solution

= - e^{4 - x} x - e^{4 - x} + C

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What is the integral of xe^{4-x} ?
The integral of xe^{4-x} is -e^{4-x}x-e^{4-x}+C