What is the integral from 0 to 0.4 of x^2(1-x)^2 ?

The integral from 0 to 0.4 of x^2(1-x)^2 is 0.01058…

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integral from 0 to 0.4 of x^2(1-x)^2

\int_{0}^{0.4} x^{2} (1 - x)^{2} d x

0.01058 \dots

Solution steps

\int_{0}^{0.4} x^{2} (1 - x)^{2} d x

Expand

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

= \int_{0}^{0.4} x^{2} - 2 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

= \int_{0}^{0.4} x^{2} d x - \int_{0}^{0.4} 2 x^{3} d x + \int_{0}^{0.4} x^{4} d x

\int_{0}^{0.4} x^{2} d x = 0.02133 \dots

\int_{0}^{0.4} 2 x^{3} d x = 0.0128

\int_{0}^{0.4} x^{4} d x = 0.002048

= 0.02133 \dots - 0.0128 + 0.002048

Simplify

= 0.01058 \dots

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What is the integral from 0 to 0.4 of x^2(1-x)^2 ?
The integral from 0 to 0.4 of x^2(1-x)^2 is 0.01058…