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

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

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integral from 0 to 1 of 6(x-x^2)

\int_{0}^{1} 6 (x - x^{2}) d x

1

Solution steps

\int_{0}^{1} 6 (x - x^{2}) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} 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}^{1} x d x - \int_{0}^{1} x^{2} d x)

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= 6 (\frac{1}{2} - \frac{1}{3})

Simplify

6 (\frac{1}{2} - \frac{1}{3}) : 1

= 1

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