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

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

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

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

\frac{π}{2}

Solution steps

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

Take the constant out:

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

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

Expand

(2 - x) (x - x^{2}) : 2 x - 3 x^{2} + x^{3}

= 2 π \cdot \int_{0}^{1} 2 x - 3 x^{2} + x^{3} 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

= 2 π (\int_{0}^{1} 2 x d x - \int_{0}^{1} 3 x^{2} d x + \int_{0}^{1} x^{3} d x)

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

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

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 2 π (1 - 1 + \frac{1}{4})

Simplify

2 π (1 - 1 + \frac{1}{4}) : \frac{π}{2}

= \frac{π}{2}

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