What is the integral from 0 to 1 of x^3(3+x^4)^3 ?

The integral from 0 to 1 of x^3(3+x^4)^3 is 175/16

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integral from 0 to 1 of x^3(3+x^4)^3

\int_{0}^{1} x^{3} (3 + x^{4})^{3} d x

\frac{175}{16}

Decimal

10.9375

Solution steps

\int_{0}^{1} 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}^{1} 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}^{1} 27 x^{3} d x + \int_{0}^{1} 27 x^{7} d x + \int_{0}^{1} 9 x^{11} d x + \int_{0}^{1} x^{15} d x

\int_{0}^{1} 27 x^{3} d x = \frac{27}{4}

\int_{0}^{1} 27 x^{7} d x = \frac{27}{8}

\int_{0}^{1} 9 x^{11} d x = \frac{3}{4}

\int_{0}^{1} x^{15} d x = \frac{1}{16}

= \frac{27}{4} + \frac{27}{8} + \frac{3}{4} + \frac{1}{16}

Simplify

\frac{27}{4} + \frac{27}{8} + \frac{3}{4} + \frac{1}{16} : \frac{175}{16}

= \frac{175}{16}

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What is the integral from 0 to 1 of x^3(3+x^4)^3 ?
The integral from 0 to 1 of x^3(3+x^4)^3 is 175/16