What is the integral of pi(2x-4)^2 ?

The integral of pi(2x-4)^2 is pi((4x^3)/3-8x^2+16x)+C

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integral of pi(2x-4)^2

\int π (2 x - 4)^{2} d x

π (\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x) + C

Solution steps

\int π (2 x - 4)^{2} d x

Take the constant out:

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

= π \cdot \int (2 x - 4)^{2} d x

Expand

(2 x - 4)^{2} : 4 x^{2} - 16 x + 16

= π \cdot \int 4 x^{2} - 16 x + 16 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 4 x^{2} d x - \int 16 x d x + \int 16 d x)

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 16 d x = 16 x

= π (\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x)

Add a constant to the solution

= π (\frac{4 x ^{3}}{3} - 8 x^{2} + 16 x) + C

ln (x - 1) + 1

\frac{\partial}{\partial z} (3 x y)

\int_{0}^{7} 8 x \cdot 3 x + 1 d x

f (x) = ln (x + 6)

\int_{0}^{15} 1000 e^{- 0.05 t} d t

What is the integral of pi(2x-4)^2 ?
The integral of pi(2x-4)^2 is pi((4x^3)/3-8x^2+16x)+C