What is the integral from 0 to 8 of pi(x+1)^2 ?

The integral from 0 to 8 of pi(x+1)^2 is pi 728/3

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integral from 0 to 8 of pi(x+1)^2

\int_{0}^{8} π (x + 1)^{2} d x

π \frac{728}{3}

Decimal

762.35981 \dots

Solution steps

\int_{0}^{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_{0}^{8} (x + 1)^{2} d x

Expand

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

= π \cdot \int_{0}^{8} x^{2} + 2 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}^{8} x^{2} d x + \int_{0}^{8} 2 x d x + \int_{0}^{8} 1 d x)

\int_{0}^{8} x^{2} d x = \frac{512}{3}

\int_{0}^{8} 2 x d x = 64

\int_{0}^{8} 1 d x = 8

= π (\frac{512}{3} + 64 + 8)

Simplify

π (\frac{512}{3} + 64 + 8) : π \frac{728}{3}

= π \frac{728}{3}

What is the integral from 0 to 8 of pi(x+1)^2 ?
The integral from 0 to 8 of pi(x+1)^2 is pi 728/3