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}

x \to 0 lim (\frac{1}{3 + x} - \frac{1}{3})

\int 6 + 7 t^{2} d t

x \to 1 + lim (\frac{( x )}{ln ( x )})

\frac{d y}{d x} = 7 x e^{4 y}

\int x (x - 1)^{8} 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