What is the integral from 0 to 9 of pi(9x-x^2) ?

The integral from 0 to 9 of pi(9x-x^2) is pi 243/2

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integral from 0 to 9 of pi(9x-x^2)

\int_{0}^{9} π (9 x - x^{2}) d x

π \frac{243}{2}

Solution steps

\int_{0}^{9} π (9 x - 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}^{9} 9 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}^{9} 9 x d x - \int_{0}^{9} x^{2} d x)

\int_{0}^{9} 9 x d x = \frac{729}{2}

\int_{0}^{9} x^{2} d x = 243

= π (\frac{729}{2} - 243)

Join

\frac{729}{2} - 243 : \frac{243}{2}

= π \frac{243}{2}

\frac{x}{3 x ^{2} + 1} - \frac{1}{8}

x \to 3 - lim (\frac{( x ^{2} )}{x - 3})

\frac{\partial}{\partial t} (\frac{t}{2})

\frac{d}{d x} (2 x^{3} - 6)

\frac{d}{d x} (24 x + sin (x))

What is the integral from 0 to 9 of pi(9x-x^2) ?
The integral from 0 to 9 of pi(9x-x^2) is pi 243/2