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

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

Popular >

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

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

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

Solution steps

\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}

= [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}}

\int_{1}^{16} (x^{- \frac{9}{4}}) d x

\int_{1}^{3} e^{- 2 x} d x

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

\int_{1}^{2} (x^{2} + 5) d x

\int_{1}^{2} 2 x + 3 d x

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