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

The integral from 0 to 1 of (x^2-x^3)^2 is 1/105

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

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

\frac{1}{105}

Decimal

0.00952 \dots

Solution steps

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

Expand

(x^{2} - x^{3})^{2} : x^{4} - 2 x^{5} + x^{6}

= \int_{0}^{1} x^{4} - 2 x^{5} + 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} x^{4} d x - \int_{0}^{1} 2 x^{5} d x + \int_{0}^{1} x^{6} d x

\int_{0}^{1} x^{4} d x = \frac{1}{5}

\int_{0}^{1} 2 x^{5} d x = \frac{1}{3}

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

= \frac{1}{5} - \frac{1}{3} + \frac{1}{7}

Simplify

\frac{1}{5} - \frac{1}{3} + \frac{1}{7} : \frac{1}{105}

= \frac{1}{105}

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What is the integral from 0 to 1 of (x^2-x^3)^2 ?
The integral from 0 to 1 of (x^2-x^3)^2 is 1/105