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

The integral from 0 to 1 of (1-x^3)^2 is 9/14

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

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

\frac{9}{14}

Decimal

0.64285 \dots

Solution steps

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

Expand

(1 - x^{3})^{2} : 1 - 2 x^{3} + x^{6}

= \int_{0}^{1} 1 - 2 x^{3} + x^{6} 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} 1 d x - \int_{0}^{1} 2 x^{3} d x + \int_{0}^{1} x^{6} d x

\int_{0}^{1} 1 d x = 1

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

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

= 1 - \frac{1}{2} + \frac{1}{7}

Simplify

1 - \frac{1}{2} + \frac{1}{7} : \frac{9}{14}

= \frac{9}{14}

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What is the integral from 0 to 1 of (1-x^3)^2 ?
The integral from 0 to 1 of (1-x^3)^2 is 9/14