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}

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\frac{d}{d x} (ln ∣ sec (x) ∣)

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

\frac{\partial}{\partial x} (2 (x^{2} + 4)^{- \frac{1}{2}})

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