What is the integral from 0 to 1 of x^{1/4}-x^{1/2} ?

The integral from 0 to 1 of x^{1/4}-x^{1/2} is 2/15

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integral from 0 to 1 of x^{1/4}-x^{1/2}

\int_{0}^{1} x^{\frac{1}{4}} - x^{\frac{1}{2}} d x

\frac{2}{15}

Decimal Notation

0.13333 \dots

Solution steps

\int_{0}^{1} x^{\frac{1}{4}} - x^{\frac{1}{2}} 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^{\frac{1}{4}} d x - \int_{0}^{1} x^{\frac{1}{2}} d x

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

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

= \frac{4}{5} - \frac{2}{3}

Simplify

\frac{4}{5} - \frac{2}{3} : \frac{2}{15}

= \frac{2}{15}

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What is the integral from 0 to 1 of x^{1/4}-x^{1/2} ?
The integral from 0 to 1 of x^{1/4}-x^{1/2} is 2/15