What is the integral from 0 to 5 of pi(sqrt(x-1))^2 ?

The integral from 0 to 5 of pi(sqrt(x-1))^2 is pi 15/2

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integral from 0 to 5 of pi(sqrt(x-1))^2

\int_{0}^{5} π (x - 1)^{2} d x

π \frac{15}{2}

Decimal Notation

23.56194 \dots

Solution steps

\int_{0}^{5} 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}^{5} (x - 1)^{2} d x

Simplify

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

= π \cdot \int_{0}^{5} 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}^{5} x d x - \int_{0}^{5} 1 d x)

\int_{0}^{5} x d x = \frac{25}{2}

\int_{0}^{5} 1 d x = 5

= π (\frac{25}{2} - 5)

Join

\frac{25}{2} - 5 : \frac{15}{2}

= π \frac{15}{2}

2 y \frac{d y}{d x} + \frac{( 1 + y ^{2} )}{x} = 1

\int sin (π) x cos (π) x d x

x \to 5 lim (- x^{2} + x - 5)

\int \frac{3 x ^{2}}{x ^{2} + 1} d x

\frac{d y}{d x} - \frac{1}{x} y = 4

What is the integral from 0 to 5 of pi(sqrt(x-1))^2 ?
The integral from 0 to 5 of pi(sqrt(x-1))^2 is pi 15/2