What is the integral from 0 to 1 of (x-6)e^x ?

The integral from 0 to 1 of (x-6)e^x is -6e+7

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integral from 0 to 1 of (x-6)e^x

\int_{0}^{1} (x - 6) e^{x} d x

- 6 e + 7

Decimal

- 9.30969 \dots

Solution steps

\int_{0}^{1} (x - 6) e^{x} d x

Apply Integration By Parts

= [e^{x} (x - 6) - \int e^{x} d x]_{0}^{1}

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

= [e^{x} (x - 6) - e^{x}]_{0}^{1}

Expand

e^{x} (x - 6) - e^{x} : e^{x} x - 7 e^{x}

= [e^{x} x - 7 e^{x}]_{0}^{1}

Compute the boundaries:

- 6 e + 7

= - 6 e + 7

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What is the integral from 0 to 1 of (x-6)e^x ?
The integral from 0 to 1 of (x-6)e^x is -6e+7