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

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

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

\int_{0}^{1} π (e - e^{x})^{2} d x

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

Decimal Notation

3.90196 \dots

Solution steps

\int_{0}^{1} π (e - e^{x})^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

(e - e^{x})^{2} : e^{2} - 2 e^{1 + x} + e^{2 x}

= π \cdot \int_{0}^{1} e^{2} - 2 e^{1 + 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

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

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

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

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

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

Simplify

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

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