What is the integral from 0 to 1 of x^{1/3}-x^3 ?

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

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integral from 0 to 1 of x^{1/3}-x^3

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

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

\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 x 9 36 + x^{2} d x

\int cos (3 x) cos (2 x) cos (x) d x

y^{^{'}} = \frac{( 6 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2} )}{x ^{3} y}

n = 0 \sum \infty \frac{1}{( n ^{2} - 1 )}

\frac{d}{d x} (ln (3 x^{2} + 1))

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