What is the integral of 3-1/3*(x-4)^2 ?

The integral of 3-1/3*(x-4)^2 is -(x^3)/9+4/3 x^2-7/3 x+C

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integral of 3-1/3*(x-4)^2

\int 3 - \frac{1}{3} \cdot (x - 4)^{2} d x

- \frac{x ^{3}}{9} + \frac{4}{3} x^{2} - \frac{7}{3} x + C

Solution steps

\int 3 - \frac{1}{3} (x - 4)^{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

= \int 3 d x - \int \frac{1}{3} (x - 4)^{2} d x

\int 3 d x = 3 x

\int \frac{1}{3} (x - 4)^{2} d x = \frac{1}{3} (\frac{x ^{3}}{3} - 4 x^{2} + 16 x)

= 3 x - \frac{1}{3} (\frac{x ^{3}}{3} - 4 x^{2} + 16 x)

Simplify

3 x - \frac{1}{3} (\frac{x ^{3}}{3} - 4 x^{2} + 16 x) : - \frac{x ^{3}}{9} + \frac{4}{3} x^{2} - \frac{7}{3} x

= - \frac{x ^{3}}{9} + \frac{4}{3} x^{2} - \frac{7}{3} x

Add a constant to the solution

= - \frac{x ^{3}}{9} + \frac{4}{3} x^{2} - \frac{7}{3} x + C

What is the integral of 3-1/3*(x-4)^2 ?
The integral of 3-1/3*(x-4)^2 is -(x^3)/9+4/3 x^2-7/3 x+C