What is the integral from 0 to 2 of 2xe^x ?

The integral from 0 to 2 of 2xe^x is 2(e^2+1)

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integral from 0 to 2 of 2xe^x

\int_{0}^{2} 2 x e^{x} d x

2 (e^{2} + 1)

Solution steps

\int_{0}^{2} 2 x e^{x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x e^{x} d x

Apply Integration By Parts

: [e^{x} x - \int e^{x} d x]_{0}^{2}

= 2 [e^{x} x - \int e^{x} d x]_{0}^{2}

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

= 2 [e^{x} x - e^{x}]_{0}^{2}

Compute the boundaries:

e^{2} + 1

= 2 (e^{2} + 1)

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What is the integral from 0 to 2 of 2xe^x ?
The integral from 0 to 2 of 2xe^x is 2(e^2+1)