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 Notation

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}

\frac{d}{d x} (- \frac{1}{2 ^{x}})

\frac{\partial}{\partial x} (1 0^{x})

\frac{x ^{2} - 2 x + 3}{x + 1}

\int \frac{1}{7} d x

\int sin (2 x) ln (sin (2 x)) d x

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