What is the integral of 6x^2(1-x) ?

The integral of 6x^2(1-x) is 6((x^3)/3-(x^4)/4)+C

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integral of 6x^2(1-x)

\int 6 x^{2} (1 - x) d x

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

Solution steps

\int 6 x^{2} (1 - x) d x

Take the constant out:

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

= 6 \cdot \int x^{2} (1 - x) d x

Expand

x^{2} (1 - x) : x^{2} - x^{3}

= 6 \cdot \int x^{2} - x^{3} 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^{3} d x)

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

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

= 6 (\frac{x ^{3}}{3} - \frac{x ^{4}}{4})

Add a constant to the solution

= 6 (\frac{x ^{3}}{3} - \frac{x ^{4}}{4}) + C

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What is the integral of 6x^2(1-x) ?
The integral of 6x^2(1-x) is 6((x^3)/3-(x^4)/4)+C