What is the integral from 1 to 8 of pi(sqrt(x-1))^2 ?

The integral from 1 to 8 of pi(sqrt(x-1))^2 is pi 49/2

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integral from 1 to 8 of pi(sqrt(x-1))^2

\int_{1}^{8} π (x - 1)^{2} d x

π \frac{49}{2}

Decimal Notation

76.96902 \dots

Solution steps

\int_{1}^{8} 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}^{8} (x - 1)^{2} d x

Simplify

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

= π \cdot \int_{1}^{8} 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}^{8} x d x - \int_{1}^{8} 1 d x)

\int_{1}^{8} x d x = \frac{63}{2}

\int_{1}^{8} 1 d x = 7

= π (\frac{63}{2} - 7)

Join

\frac{63}{2} - 7 : \frac{49}{2}

= π \frac{49}{2}

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What is the integral from 1 to 8 of pi(sqrt(x-1))^2 ?
The integral from 1 to 8 of pi(sqrt(x-1))^2 is pi 49/2