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

The integral from 0 to 1 of (x^4-x^3) is -1/20

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

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

- \frac{1}{20}

Decimal

- 0.05

Solution steps

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

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

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

= \frac{1}{5} - \frac{1}{4}

Simplify

\frac{1}{5} - \frac{1}{4} : - \frac{1}{20}

= - \frac{1}{20}

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What is the integral from 0 to 1 of (x^4-x^3) ?
The integral from 0 to 1 of (x^4-x^3) is -1/20