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

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

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Expand

x^{3} (1 - 3 x^{4})^{2} : x^{3} - 6 x^{7} + 9 x^{11}

= \int_{0}^{1} x^{3} - 6 x^{7} + 9 x^{11} 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^{3} d x - \int_{0}^{1} 6 x^{7} d x + \int_{0}^{1} 9 x^{11} d x

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

\int_{0}^{1} 6 x^{7} d x = \frac{3}{4}

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

= \frac{1}{4} - \frac{3}{4} + \frac{3}{4}

Simplify

= \frac{1}{4}

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