What is the integral from 0 to 2 of x(16-x^4)^5 ?

The integral from 0 to 2 of x(16-x^4)^5 is 536870912/693

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integral from 0 to 2 of x(16-x^4)^5

\int_{0}^{2} x (16 - x^{4})^{5} d x

\frac{536870912}{693}

Decimal

774705.50072 \dots

Solution steps

\int_{0}^{2} x (16 - x^{4})^{5} d x

Expand

x (16 - x^{4})^{5} : 1048576 x - 327680 x^{5} + 40960 x^{9} - 2560 x^{13} + 80 x^{17} - x^{21}

= \int_{0}^{2} 1048576 x - 327680 x^{5} + 40960 x^{9} - 2560 x^{13} + 80 x^{17} - x^{21} 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}^{2} 1048576 x d x - \int_{0}^{2} 327680 x^{5} d x + \int_{0}^{2} 40960 x^{9} d x - \int_{0}^{2} 2560 x^{13} d x + \int_{0}^{2} 80 x^{17} d x - \int_{0}^{2} x^{21} d x

\int_{0}^{2} 1048576 x d x = 2097152

\int_{0}^{2} 327680 x^{5} d x = \frac{10485760}{3}

\int_{0}^{2} 40960 x^{9} d x = 4194304

\int_{0}^{2} 2560 x^{13} d x = \frac{20971520}{7}

\int_{0}^{2} 80 x^{17} d x = \frac{10485760}{9}

\int_{0}^{2} x^{21} d x = \frac{2097152}{11}

= 2097152 - \frac{10485760}{3} + 4194304 - \frac{20971520}{7} + \frac{10485760}{9} - \frac{2097152}{11}

Simplify

2097152 - \frac{10485760}{3} + 4194304 - \frac{20971520}{7} + \frac{10485760}{9} - \frac{2097152}{11} : \frac{536870912}{693}

= \frac{536870912}{693}

What is the integral from 0 to 2 of x(16-x^4)^5 ?
The integral from 0 to 2 of x(16-x^4)^5 is 536870912/693