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

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

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

\int_{0}^{1} π (1 - x)^{2} d x

\frac{π}{3}

Decimal Notation

1.04719 \dots

Solution steps

\int_{0}^{1} π (1 - 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} (1 - x)^{2} d x

Expand

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

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

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

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

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

= π (1 - 1 + \frac{1}{3})

Simplify

π (1 - 1 + \frac{1}{3}) : \frac{π}{3}

= \frac{π}{3}

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