What is the integral from 0 to 1 of [sqrt(x)-x^2] ?

The integral from 0 to 1 of [sqrt(x)-x^2] is 1/3

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integral from 0 to 1 of [sqrt(x)-x^2]

\int_{0}^{1} [x - x^{2}] d x

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

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

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

= \frac{2}{3} - \frac{1}{3}

Simplify

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

= \frac{1}{3}

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