What is the integral from 0 to 1 of x^2(5x^3-1)^4 ?

The integral from 0 to 1 of x^2(5x^3-1)^4 is 41/3

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integral from 0 to 1 of x^2(5x^3-1)^4

\int_{0}^{1} x^{2} (5 x^{3} - 1)^{4} d x

\frac{41}{3}

Decimal Notation

13.66666 \dots

Solution steps

\int_{0}^{1} x^{2} (5 x^{3} - 1)^{4} d x

Expand

x^{2} (5 x^{3} - 1)^{4} : 625 x^{14} - 500 x^{11} + 150 x^{8} - 20 x^{5} + x^{2}

= \int_{0}^{1} 625 x^{14} - 500 x^{11} + 150 x^{8} - 20 x^{5} + x^{2} 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} 625 x^{14} d x - \int_{0}^{1} 500 x^{11} d x + \int_{0}^{1} 150 x^{8} d x - \int_{0}^{1} 20 x^{5} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} 625 x^{14} d x = \frac{125}{3}

\int_{0}^{1} 500 x^{11} d x = \frac{125}{3}

\int_{0}^{1} 150 x^{8} d x = \frac{50}{3}

\int_{0}^{1} 20 x^{5} d x = \frac{10}{3}

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

= \frac{125}{3} - \frac{125}{3} + \frac{50}{3} - \frac{10}{3} + \frac{1}{3}

Simplify

\frac{125}{3} - \frac{125}{3} + \frac{50}{3} - \frac{10}{3} + \frac{1}{3} : \frac{41}{3}

= \frac{41}{3}

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What is the integral from 0 to 1 of x^2(5x^3-1)^4 ?
The integral from 0 to 1 of x^2(5x^3-1)^4 is 41/3