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

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

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

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

\frac{1}{6}

Decimal Notation

0.16666 \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{1}{2}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

Simplify

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

= \frac{1}{6}

\int (x - 7)^{2} d x

n = 1 \sum \infty \frac{1}{2 ^{n - 1}}

x \to \infty lim (0^{0})

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

\int_{0}^{1} 2 x^{2} (x^{3} - 1)^{3} d x

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