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

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

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

\int_{0}^{1} e^{x} (e^{x} - 1) d x

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

Decimal

1.47624 \dots

Solution steps

\int_{0}^{1} e^{x} (e^{x} - 1) d x

Apply u-substitution

: \int_{1}^{e} u - 1 d u

= \int_{1}^{e} u - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{e} u d u - \int_{1}^{e} 1 d u

\int_{1}^{e} u d u = \frac{e ^{2} - 1}{2}

\int_{1}^{e} 1 d u = e - 1

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

Simplify

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

\int e^{x} (- x) d x

\int \frac{1}{u ln ( u ) - u} d u

30 t^{2}

\frac{d}{d x} (\frac{1}{1 - 2 x})

x \to \infty lim (- ln (x))

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