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

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

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integral from 0 to 2 of (xe^x)

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

e^{2} + 1

Decimal Notation

8.38905 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

e^{2} + 1

= e^{2} + 1

\int_{2}^{e} \frac{1}{x ( ln ( x ) ) ^{2}} d x

\int_{- 1}^{1} sec^{2} (x) d x

\int_{- 1}^{1} 1 + x^{3} d x

\int_{0}^{2} x^{2} - 3 x d x

\int_{0}^{4} (2 x + 1)^{- \frac{1}{2}} d x

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