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

The integral from 0 to 1 of (x^4-x^5) is 1/30

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

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

\frac{1}{30}

Solution steps

\int_{0}^{1} x^{4} - x^{5} 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} x^{5} d x

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

\int_{0}^{1} x^{5} d x = \frac{1}{6}

= \frac{1}{5} - \frac{1}{6}

Simplify

\frac{1}{5} - \frac{1}{6} : \frac{1}{30}

= \frac{1}{30}

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What is the integral from 0 to 1 of (x^4-x^5) ?
The integral from 0 to 1 of (x^4-x^5) is 1/30