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

The integral from 0 to t of xe^x is te^t-e^t+1

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integral from 0 to t of xe^x

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

t e^{t} - e^{t} + 1

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

t e^{t} - e^{t} + 1

= t e^{t} - e^{t} + 1

\int_{0}^{1} 1 + t^{2} d t

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

\int_{2 x}^{x^{2}} e^{- t^{2}} d t

\int_{- 2}^{4} 6 + 4 x - x^{2} d x

\int_{1}^{e - 1} ln (x + 1) d x

What is the integral from 0 to t of xe^x ?
The integral from 0 to t of xe^x is te^t-e^t+1