What is the integral from 1 to 9 of pi(sqrt(x-1))^2 ?

The integral from 1 to 9 of pi(sqrt(x-1))^2 is 32pi

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integral from 1 to 9 of pi(sqrt(x-1))^2

\int_{1}^{9} π (x - 1)^{2} d x

32 π

Decimal Notation

100.53096 \dots

Solution steps

\int_{1}^{9} π (x - 1)^{2} d x

Take the constant out:

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

= π \cdot \int_{1}^{9} (x - 1)^{2} d x

Simplify

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

= π \cdot \int_{1}^{9} x - 1 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_{1}^{9} x d x - \int_{1}^{9} 1 d x)

\int_{1}^{9} x d x = 40

\int_{1}^{9} 1 d x = 8

= π (40 - 8)

Simplify

= 32 π

s \to 0 lim (\frac{3}{s \cdot ( s + 2 )})

\frac{x ^{3}}{9}

(sin (x y) + x y cos (x y)) d x + (x^{2} cos (x y)) d y = 0

x \to 2 + lim (\frac{x ^{2} + 1}{x - 2})

\int_{1}^{\infty} e^{- 11 x} d x

What is the integral from 1 to 9 of pi(sqrt(x-1))^2 ?
The integral from 1 to 9 of pi(sqrt(x-1))^2 is 32pi