What is the integral from 0 to 1/2 of 60x^2(1-x)^3 ?

The integral from 0 to 1/2 of 60x^2(1-x)^3 is 21/32

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integral from 0 to 1/2 of 60x^2(1-x)^3

\int_{0}^{\frac{1}{2}} 60 x^{2} (1 - x)^{3} d x

\frac{21}{32}

Decimal

0.65625

Solution steps

\int_{0}^{\frac{1}{2}} 60 x^{2} (1 - x)^{3} d x

Take the constant out:

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

= 60 \cdot \int_{0}^{\frac{1}{2}} x^{2} (1 - x)^{3} d x

Expand

x^{2} (1 - x)^{3} : x^{2} - 3 x^{3} + 3 x^{4} - x^{5}

= 60 \cdot \int_{0}^{\frac{1}{2}} x^{2} - 3 x^{3} + 3 x^{4} - x^{5} 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

= 60 (\int_{0}^{\frac{1}{2}} x^{2} d x - \int_{0}^{\frac{1}{2}} 3 x^{3} d x + \int_{0}^{\frac{1}{2}} 3 x^{4} d x - \int_{0}^{\frac{1}{2}} x^{5} d x)

\int_{0}^{\frac{1}{2}} x^{2} d x = \frac{1}{24}

\int_{0}^{\frac{1}{2}} 3 x^{3} d x = \frac{3}{64}

\int_{0}^{\frac{1}{2}} 3 x^{4} d x = \frac{3}{160}

\int_{0}^{\frac{1}{2}} x^{5} d x = \frac{1}{384}

= 60 (\frac{1}{24} - \frac{3}{64} + \frac{3}{160} - \frac{1}{384})

Simplify

60 (\frac{1}{24} - \frac{3}{64} + \frac{3}{160} - \frac{1}{384}) : \frac{21}{32}

= \frac{21}{32}

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What is the integral from 0 to 1/2 of 60x^2(1-x)^3 ?
The integral from 0 to 1/2 of 60x^2(1-x)^3 is 21/32