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

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

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integral of 2(x-4)^2

\int 2 (x - 4)^{2} d x

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

Solution steps

\int 2 (x - 4)^{2} d x

Take the constant out:

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

= 2 \cdot \int (x - 4)^{2} d x

Expand

(x - 4)^{2} : x^{2} - 8 x + 16

= 2 \cdot \int x^{2} - 8 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

= 2 (\int x^{2} d x - \int 8 x d x + \int 16 d x)

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

\int 8 x d x = 4 x^{2}

\int 16 d x = 16 x

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

Add a constant to the solution

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

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