What is the integral from 0 to 1 of 17xe^x ?

The integral from 0 to 1 of 17xe^x is 17

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integral from 0 to 1 of 17xe^x

\int_{0}^{1} 17 x e^{x} d x

17

Solution steps

\int_{0}^{1} 17 x e^{x} d x

Take the constant out:

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

= 17 \cdot \int_{0}^{1} x e^{x} d x

Apply Integration By Parts

: [e^{x} x - \int e^{x} d x]_{0}^{1}

= 17 [e^{x} x - \int e^{x} d x]_{0}^{1}

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

= 17 [e^{x} x - e^{x}]_{0}^{1}

Compute the boundaries:

1

= 17 \cdot 1

Simplify

= 17

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What is the integral from 0 to 1 of 17xe^x ?
The integral from 0 to 1 of 17xe^x is 17