What is the integral from 0 to 1 of x(2-x^2)^5 ?

The integral from 0 to 1 of x(2-x^2)^5 is 21/4

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integral from 0 to 1 of x(2-x^2)^5

\int_{0}^{1} x (2 - x^{2})^{5} d x

\frac{21}{4}

Decimal

5.25

Solution steps

\int_{0}^{1} x (2 - x^{2})^{5} d x

Expand

x (2 - x^{2})^{5} : 32 x - 80 x^{3} + 80 x^{5} - 40 x^{7} + 10 x^{9} - x^{11}

= \int_{0}^{1} 32 x - 80 x^{3} + 80 x^{5} - 40 x^{7} + 10 x^{9} - x^{11} 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} 32 x d x - \int_{0}^{1} 80 x^{3} d x + \int_{0}^{1} 80 x^{5} d x - \int_{0}^{1} 40 x^{7} d x + \int_{0}^{1} 10 x^{9} d x - \int_{0}^{1} x^{11} d x

\int_{0}^{1} 32 x d x = 16

\int_{0}^{1} 80 x^{3} d x = 20

\int_{0}^{1} 80 x^{5} d x = \frac{40}{3}

\int_{0}^{1} 40 x^{7} d x = 5

\int_{0}^{1} 10 x^{9} d x = 1

\int_{0}^{1} x^{11} d x = \frac{1}{12}

= 16 - 20 + \frac{40}{3} - 5 + 1 - \frac{1}{12}

Simplify

16 - 20 + \frac{40}{3} - 5 + 1 - \frac{1}{12} : \frac{21}{4}

= \frac{21}{4}

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What is the integral from 0 to 1 of x(2-x^2)^5 ?
The integral from 0 to 1 of x(2-x^2)^5 is 21/4