What is the integral from 0 to 6 of pi(sqrt(x-1))^2 ?

The integral from 0 to 6 of pi(sqrt(x-1))^2 is 12pi

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integral from 0 to 6 of pi(sqrt(x-1))^2

\int_{0}^{6} π (x - 1)^{2} d x

12 π

Decimal Notation

37.69911 \dots

Solution steps

\int_{0}^{6} 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_{0}^{6} (x - 1)^{2} d x

Simplify

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

= π \cdot \int_{0}^{6} 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_{0}^{6} x d x - \int_{0}^{6} 1 d x)

\int_{0}^{6} x d x = 18

\int_{0}^{6} 1 d x = 6

= π (18 - 6)

Simplify

= 12 π

\int \frac{1}{1 - x ^{2}} d x

x^{2} y^{^{'}} + 8 x y = 17 y^{3}

\int \frac{2 x arcsin ( x )}{1 - x ^{2}} d x

\frac{d}{d x} (1 + ln (x))

\frac{d}{d x} ((2 x^{3} - 3 x + 1) (x^{2} + 5 x))

What is the integral from 0 to 6 of pi(sqrt(x-1))^2 ?
The integral from 0 to 6 of pi(sqrt(x-1))^2 is 12pi