What is the integral from 0 to x of 6x(1-x) ?

The integral from 0 to x of 6x(1-x) is 6((x^2)/2-(x^3)/3)

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integral from 0 to x of 6x(1-x)

\int_{0}^{x} 6 x (1 - x) d x

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

Solution steps

\int_{0}^{x} 6 x (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_{0}^{x} x (1 - x) d x

Expand

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

= 6 \cdot \int_{0}^{x} x - 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

= 6 (\int_{0}^{x} x d x - \int_{0}^{x} x^{2} d x)

\int_{0}^{x} x d x = \frac{x ^{2}}{2}

\int_{0}^{x} x^{2} d x = \frac{x ^{3}}{3}

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

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What is the integral from 0 to x of 6x(1-x) ?
The integral from 0 to x of 6x(1-x) is 6((x^2)/2-(x^3)/3)