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

The integral from 0 to 1 of (x-e^{-x})^2 is -5/3+4/e+(e^2-1)/(2e^2)

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

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

- \frac{5}{3} + \frac{4}{e} + \frac{e ^{2} - 1}{2 e ^{2}}

Decimal

0.23718 \dots

Solution steps

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

Apply u-substitution

= \int_{0}^{- 1} - u^{2} - 2 e^{u} u - e^{2 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} - u^{2} - 2 e^{u} u - e^{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}^{0} u^{2} d u - \int_{- 1}^{0} 2 e^{u} u d u - \int_{- 1}^{0} e^{2 u} d u)

\int_{- 1}^{0} u^{2} d u = \frac{1}{3}

\int_{- 1}^{0} 2 e^{u} u d u = 2 (- 1 + \frac{2}{e})

\int_{- 1}^{0} e^{2 u} d u = \frac{e ^{2} - 1}{2 e ^{2}}

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

Simplify

- (- \frac{1}{3} - 2 (- 1 + \frac{2}{e}) - \frac{e ^{2} - 1}{2 e ^{2}}) : - \frac{5}{3} + \frac{4}{e} + \frac{e ^{2} - 1}{2 e ^{2}}

= - \frac{5}{3} + \frac{4}{e} + \frac{e ^{2} - 1}{2 e ^{2}}

What is the integral from 0 to 1 of (x-e^{-x})^2 ?
The integral from 0 to 1 of (x-e^{-x})^2 is -5/3+4/e+(e^2-1)/(2e^2)