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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Add a constant to the solution

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

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