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+(e^2-1)/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{e ^{2} - 1}{2}

Decimal Notation

1.52786 \dots

Solution steps

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

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

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

Simplify

((x + e^{\frac{x}{2}}) (x - e^{\frac{x}{2}}))^{2} : x^{2} - 2 e^{x} x + e^{2 x}

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

Apply the Sum Rule:

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

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

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

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

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

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

Simplify

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

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

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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+(e^2-1)/2