What is the integral from 0 to 6 of pi/8 (6-x)^2 ?

The integral from 0 to 6 of pi/8 (6-x)^2 is 9pi

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integral from 0 to 6 of pi/8 (6-x)^2

\int_{0}^{6} \frac{π}{8} (6 - x)^{2} d x

9 π

Decimal

28.27433 \dots

Solution steps

\int_{0}^{6} \frac{π}{8} (6 - x)^{2} d x

Take the constant out:

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

= \frac{π}{8} \cdot \int_{0}^{6} (6 - x)^{2} d x

Expand

(6 - x)^{2} : 36 - 12 x + x^{2}

= \frac{π}{8} \cdot \int_{0}^{6} 36 - 12 x + x^{2} 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

= \frac{π}{8} (\int_{0}^{6} 36 d x - \int_{0}^{6} 12 x d x + \int_{0}^{6} x^{2} d x)

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

\int_{0}^{6} 12 x d x = 216

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

= \frac{π}{8} (216 - 216 + 72)

Simplify

\frac{π}{8} (216 - 216 + 72) : 9 π

= 9 π

\int sin^{4} (x co) s^{2} x d x

y = \frac{cos ( x )}{x ^{7}}

x \to 0 lim (\frac{sin ( 2 x )}{x})

y^{^{'}} = - y^{3}, y (0) = 2

\frac{\partial}{\partial x} (7 x cos (x y))

What is the integral from 0 to 6 of pi/8 (6-x)^2 ?
The integral from 0 to 6 of pi/8 (6-x)^2 is 9pi