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

The integral from 0 to 1 of 6x^3(1-x) is 3/10

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integral from 0 to 1 of 6x^3(1-x)

\int_{0}^{1} 6 x^{3} (1 - x) d x

\frac{3}{10}

Decimal

0.3

Solution steps

\int_{0}^{1} 6 x^{3} (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^{3} (1 - x) d x

Expand

x^{3} (1 - x) : x^{3} - x^{4}

= 6 \cdot \int_{0}^{1} x^{3} - x^{4} 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^{3} d x - \int_{0}^{1} x^{4} d x)

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

\int_{0}^{1} x^{4} d x = \frac{1}{5}

= 6 (\frac{1}{4} - \frac{1}{5})

Simplify

6 (\frac{1}{4} - \frac{1}{5}) : \frac{3}{10}

= \frac{3}{10}

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What is the integral from 0 to 1 of 6x^3(1-x) ?
The integral from 0 to 1 of 6x^3(1-x) is 3/10