What is the integral from 1 to e^4 of (5-x)e^x ?

The integral from 1 to e^4 of (5-x)e^x is -e^{e^4+4}+6e^{e^4}-5e

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integral from 1 to e^4 of (5-x)e^x

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

- e^{e^{4} + 4} + 6 e^{e^{4}} - 5 e

Solution steps

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

Apply Integration By Parts

: [e^{x} (5 - x) - \int - e^{x} d x]_{1}^{e^{4}}

= [e^{x} (5 - x) - \int - e^{x} d x]_{1}^{e^{4}}

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

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

Simplify

[e^{x} (5 - x) - (- e^{x})]_{1}^{e^{4}} : [- e^{x} x + 6 e^{x}]_{1}^{e^{4}}

= [- e^{x} x + 6 e^{x}]_{1}^{e^{4}}

Compute the boundaries:

- e^{e^{4} + 4} + 6 e^{e^{4}} - 5 e

= - e^{e^{4} + 4} + 6 e^{e^{4}} - 5 e

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What is the integral from 1 to e^4 of (5-x)e^x ?
The integral from 1 to e^4 of (5-x)e^x is -e^{e^4+4}+6e^{e^4}-5e