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

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

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

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

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

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

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

Simplify

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

= \frac{1}{3}

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