What is the integral from 0 to 1 of-(x^2-x) ?

The integral from 0 to 1 of-(x^2-x) is 1/6

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integral from 0 to 1 of-(x^2-x)

\int_{0}^{1} - (x^{2} - x) d x

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

\int_{0}^{1} - (x^{2} - x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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{1}{2}

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

Simplify

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

= \frac{1}{6}

x \to 3 lim (- 7 - x)

\int x e^{2 x} d x

y^{^{''''}} - 4 y^{^{''}} = 0

\frac{d}{d x} (- \frac{2 x ^{2}}{3 y ^{2}})

\int - 9 sin (x) d x

What is the integral from 0 to 1 of-(x^2-x) ?
The integral from 0 to 1 of-(x^2-x) is 1/6