What is the integral from 0 to 1 of \sqrt[3]{x}-x^3 ?

The integral from 0 to 1 of \sqrt[3]{x}-x^3 is 1/2

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integral from 0 to 1 of \sqrt[3]{x}-x^3

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

= \frac{3}{4} - \frac{1}{4}

Simplify

\frac{3}{4} - \frac{1}{4} : \frac{1}{2}

= \frac{1}{2}

\int_{0}^{1} sin (π t) d t

\frac{d}{d x} (\frac{x}{x ^{2} + y ^{2}})

\int 1 - cos^{2} (x) d x

\frac{d}{d x} (- 4 x^{3} + 4 x)

x^{2}, (0, 3)

What is the integral from 0 to 1 of \sqrt[3]{x}-x^3 ?
The integral from 0 to 1 of \sqrt[3]{x}-x^3 is 1/2