What is the integral of 6x(x-1) ?

The integral of 6x(x-1) is 6((x^3)/3-(x^2)/2)+C

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integral of 6x(x-1)

\int 6 x (x - 1) d x

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

Solution steps

\int 6 x (x - 1) d x

Take the constant out:

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

= 6 \cdot \int x (x - 1) d x

Expand

x (x - 1) : x^{2} - x

= 6 \cdot \int x^{2} - x 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

= 6 (\int x^{2} d x - \int x d x)

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

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

= 6 (\frac{x ^{3}}{3} - \frac{x ^{2}}{2})

Add a constant to the solution

= 6 (\frac{x ^{3}}{3} - \frac{x ^{2}}{2}) + C

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What is the integral of 6x(x-1) ?
The integral of 6x(x-1) is 6((x^3)/3-(x^2)/2)+C