What is the integral from 0 to 1 of x^2(3.6x-2.4x^2) ?

The integral from 0 to 1 of x^2(3.6x-2.4x^2) is 0.42

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integral from 0 to 1 of x^2(3.6x-2.4x^2)

\int_{0}^{1} x^{2} (3.6 x - 2.4 x^{2}) d x

0.42

Solution steps

\int_{0}^{1} x^{2} (3.6 x - 2.4 x^{2}) d x

Expand

x^{2} (3.6 x - 2.4 x^{2}) : 3.6 x^{3} - 2.4 x^{4}

= \int_{0}^{1} 3.6 x^{3} - 2.4 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}^{1} 3.6 x^{3} d x - \int_{0}^{1} 2.4 x^{4} d x

\int_{0}^{1} 3.6 x^{3} d x = 0.9

\int_{0}^{1} 2.4 x^{4} d x = 0.48

= 0.9 - 0.48

Simplify

= 0.42

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What is the integral from 0 to 1 of x^2(3.6x-2.4x^2) ?
The integral from 0 to 1 of x^2(3.6x-2.4x^2) is 0.42