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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Expand

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

= 6 \cdot \int_{0}^{1} 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_{0}^{1} x^{2} d x - \int_{0}^{1} x^{3} d x)

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

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 6 (\frac{1}{3} - \frac{1}{4})

Simplify

6 (\frac{1}{3} - \frac{1}{4}) : \frac{1}{2}

= \frac{1}{2}

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