What is the integral from 0 to 12 of pi(1+x/3)^2 ?

The integral from 0 to 12 of pi(1+x/3)^2 is 124pi

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integral from 0 to 12 of pi(1+x/3)^2

\int_{0}^{12} π (1 + \frac{x}{3})^{2} d x

124 π

Decimal

389.55748 \dots

Solution steps

\int_{0}^{12} π (1 + \frac{x}{3})^{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}^{12} (1 + \frac{x}{3})^{2} d x

Expand

(1 + \frac{x}{3})^{2} : 1 + \frac{2 x}{3} + \frac{x ^{2}}{9}

= π \cdot \int_{0}^{12} 1 + \frac{2 x}{3} + \frac{x ^{2}}{9} 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}^{12} 1 d x + \int_{0}^{12} \frac{2 x}{3} d x + \int_{0}^{12} \frac{x ^{2}}{9} d x)

\int_{0}^{12} 1 d x = 12

\int_{0}^{12} \frac{2 x}{3} d x = 48

\int_{0}^{12} \frac{x ^{2}}{9} d x = 64

= π (12 + 48 + 64)

Simplify

= 124 π

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What is the integral from 0 to 12 of pi(1+x/3)^2 ?
The integral from 0 to 12 of pi(1+x/3)^2 is 124pi