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

The integral from 0 to t of (t-x)e^x is e^t-t-1

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integral from 0 to t of (t-x)e^x

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

e^{t} - t - 1

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

e^{t} - t - 1

= e^{t} - t - 1

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