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

The integral from 0 to 1 of (1-e^x)^2 is 3+(e^2-1)/2-2e

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integral from 0 to 1 of (1-e^x)^2

\int_{0}^{1} (1 - e^{x})^{2} d x

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

Decimal

0.75796 \dots

Solution steps

\int_{0}^{1} (1 - e^{x})^{2} d x

Apply u-substitution

: \int_{1}^{e} \frac{( 1 - u ) ^{2}}{u} d u

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

Expand

\frac{( 1 - u ) ^{2}}{u} : \frac{1}{u} - 2 + u

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

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

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

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

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

Simplify

= 3 + \frac{e ^{2} - 1}{2} - 2 e

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What is the integral from 0 to 1 of (1-e^x)^2 ?
The integral from 0 to 1 of (1-e^x)^2 is 3+(e^2-1)/2-2e