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 Notation

- 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}

- 2 x^{5} + 4 x^{4}

\frac{d}{d x} (\frac{12}{x} - \frac{4}{x ^{3}} + \frac{1}{x ^{4}})

y^{^{''}} + 2 y^{^{'}} + y = 0, y (2) = 1, y^{^{'}} (2) = - 2

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

\frac{d}{d x} (11 x - \frac{6}{x})

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