What is the integral from 1 to 7 of pi(sqrt(x-1))^2 ?

The integral from 1 to 7 of pi(sqrt(x-1))^2 is 18pi

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integral from 1 to 7 of pi(sqrt(x-1))^2

\int_{1}^{7} π (x - 1)^{2} d x

18 π

Decimal Notation

56.54866 \dots

Solution steps

\int_{1}^{7} 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}^{7} (x - 1)^{2} d x

Simplify

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

= π \cdot \int_{1}^{7} 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}^{7} x d x - \int_{1}^{7} 1 d x)

\int_{1}^{7} x d x = 24

\int_{1}^{7} 1 d x = 6

= π (24 - 6)

Simplify

= 18 π

y = 3 x^{\frac{2}{3}} - 2 x

\int_{- 3}^{0} (x^{2} - 4 x + 7) d x

\int (x + 3) 2 - x d x

x \to 4 - lim (5 (x) - 7)

arccos (1 - x^{2})

What is the integral from 1 to 7 of pi(sqrt(x-1))^2 ?
The integral from 1 to 7 of pi(sqrt(x-1))^2 is 18pi