What is the integral from 0 to 1 of xe^{tx} ?

The integral from 0 to 1 of xe^{tx} is (e^t)/t+(-e^t+1)/(t^2)

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integral from 0 to 1 of xe^{tx}

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

\frac{e ^{t}}{t} + \frac{- e ^{t} + 1}{t ^{2}}

Solution steps

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

Apply Integration By Parts

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

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

\int e^{t x} \frac{1}{t} d x = e^{t x} \frac{1}{t ^{2}}

= [e^{t x} \frac{1}{t} x - e^{t x} \frac{1}{t ^{2}}]_{0}^{1}

Compute the boundaries:

e^{t} \frac{1}{t} - e^{t} \frac{1}{t ^{2}} + \frac{1}{t ^{2}}

= e^{t} \frac{1}{t} - e^{t} \frac{1}{t ^{2}} + \frac{1}{t ^{2}}

Simplify

= \frac{e ^{t}}{t} + \frac{- e ^{t} + 1}{t ^{2}}

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What is the integral from 0 to 1 of xe^{tx} ?
The integral from 0 to 1 of xe^{tx} is (e^t)/t+(-e^t+1)/(t^2)