What is the integral from 0 to 1 of x^2(1+3x^3)^5 ?

The integral from 0 to 1 of x^2(1+3x^3)^5 is 455/6

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integral from 0 to 1 of x^2(1+3x^3)^5

\int_{0}^{1} x^{2} (1 + 3 x^{3})^{5} d x

\frac{455}{6}

Decimal

75.83333 \dots

Solution steps

\int_{0}^{1} x^{2} (1 + 3 x^{3})^{5} d x

Expand

x^{2} (1 + 3 x^{3})^{5} : x^{2} + 15 x^{5} + 90 x^{8} + 270 x^{11} + 405 x^{14} + 243 x^{17}

= \int_{0}^{1} x^{2} + 15 x^{5} + 90 x^{8} + 270 x^{11} + 405 x^{14} + 243 x^{17} 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} x^{2} d x + \int_{0}^{1} 15 x^{5} d x + \int_{0}^{1} 90 x^{8} d x + \int_{0}^{1} 270 x^{11} d x + \int_{0}^{1} 405 x^{14} d x + \int_{0}^{1} 243 x^{17} d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 15 x^{5} d x = \frac{5}{2}

\int_{0}^{1} 90 x^{8} d x = 10

\int_{0}^{1} 270 x^{11} d x = \frac{45}{2}

\int_{0}^{1} 405 x^{14} d x = 27

\int_{0}^{1} 243 x^{17} d x = \frac{27}{2}

= \frac{1}{3} + \frac{5}{2} + 10 + \frac{45}{2} + 27 + \frac{27}{2}

Simplify

\frac{1}{3} + \frac{5}{2} + 10 + \frac{45}{2} + 27 + \frac{27}{2} : \frac{455}{6}

= \frac{455}{6}

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What is the integral from 0 to 1 of x^2(1+3x^3)^5 ?
The integral from 0 to 1 of x^2(1+3x^3)^5 is 455/6