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

The integral from 0 to 1 of (sqrt(x)-x) is 1/6

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integral from 0 to 1 of (sqrt(x)-x)

\int_{0}^{1} (x - x) d x

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

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

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

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

Simplify

\frac{2}{3} - \frac{1}{2} : \frac{1}{6}

= \frac{1}{6}

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What is the integral from 0 to 1 of (sqrt(x)-x) ?
The integral from 0 to 1 of (sqrt(x)-x) is 1/6