What is the integral from 0 to 1 of e^{tx}2x ?

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

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integral from 0 to 1 of e^{tx}2x

\int_{0}^{1} e^{t x} 2 x d x

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

Solution steps

\int_{0}^{1} e^{t x} \cdot 2 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}^{1} e^{t x} 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}

= 2 [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}}

= 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}}

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

Simplify

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

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