What is the integral from 0 to 2 of x^3(3+x^4)^3 ?

The integral from 0 to 2 of x^3(3+x^4)^3 is 8140

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integral from 0 to 2 of x^3(3+x^4)^3

\int_{0}^{2} x^{3} (3 + x^{4})^{3} d x

8140

Solution steps

\int_{0}^{2} x^{3} (3 + x^{4})^{3} d x

Expand

x^{3} (3 + x^{4})^{3} : 27 x^{3} + 27 x^{7} + 9 x^{11} + x^{15}

= \int_{0}^{2} 27 x^{3} + 27 x^{7} + 9 x^{11} + x^{15} 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} 27 x^{3} d x + \int_{0}^{2} 27 x^{7} d x + \int_{0}^{2} 9 x^{11} d x + \int_{0}^{2} x^{15} d x

\int_{0}^{2} 27 x^{3} d x = 108

\int_{0}^{2} 27 x^{7} d x = 864

\int_{0}^{2} 9 x^{11} d x = 3072

\int_{0}^{2} x^{15} d x = 4096

= 108 + 864 + 3072 + 4096

Simplify

= 8140

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What is the integral from 0 to 2 of x^3(3+x^4)^3 ?
The integral from 0 to 2 of x^3(3+x^4)^3 is 8140