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 2/3

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

\int_{0}^{1} π (1 - x^{2}) d x

π \frac{2}{3}

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

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} x^{2} d x)

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

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

= π (1 - \frac{1}{3})

Join

1 - \frac{1}{3} : \frac{2}{3}

= π \frac{2}{3}

\frac{- 5}{s ( s ^{2} - 5 s + 4 )}

y = - 3 x - 2

x^{^{''}} + 6 x^{^{'}} + 7 x = 0, x (0) = 0, x^{^{'}} (0) = 1

\int_{2}^{5} (4 x - 2) d x

(x)^{^{'}}

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 2/3