What is the integral from 0 to 6 of pi/9 (36-y^2) ?

The integral from 0 to 6 of pi/9 (36-y^2) is 16pi

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integral from 0 to 6 of pi/9 (36-y^2)

\int_{0}^{6} \frac{π}{9} (36 - y^{2}) d y

16 π

Decimal Notation

50.26548 \dots

Solution steps

\int_{0}^{6} \frac{π}{9} (36 - y^{2}) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{π}{9} \cdot \int_{0}^{6} 36 - y^{2} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{π}{9} (\int_{0}^{6} 36 d y - \int_{0}^{6} y^{2} d y)

\int_{0}^{6} 36 d y = 216

\int_{0}^{6} y^{2} d y = 72

= \frac{π}{9} (216 - 72)

Simplify

\frac{π}{9} (216 - 72) : 16 π

= 16 π

(sin (x) - x cos (x))^{^{'}}

y^{^{'}} = 20 x y

\int 3 x sin (\frac{1}{6} x) d x

f (x) = - \frac{1}{x}

n = 1 \sum \infty 6 sin (\frac{1}{n})

What is the integral from 0 to 6 of pi/9 (36-y^2) ?
The integral from 0 to 6 of pi/9 (36-y^2) is 16pi