What is the integral from 0 to 1/9 of x-9x^2 ?

The integral from 0 to 1/9 of x-9x^2 is 1/486

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integral from 0 to 1/9 of x-9x^2

\int_{0}^{\frac{1}{9}} x - 9 x^{2} d x

\frac{1}{486}

Decimal Notation

0.00205 \dots

Solution steps

\int_{0}^{\frac{1}{9}} x - 9 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}^{\frac{1}{9}} x d x - \int_{0}^{\frac{1}{9}} 9 x^{2} d x

\int_{0}^{\frac{1}{9}} x d x = \frac{1}{162}

\int_{0}^{\frac{1}{9}} 9 x^{2} d x = \frac{1}{243}

= \frac{1}{162} - \frac{1}{243}

Simplify

\frac{1}{162} - \frac{1}{243} : \frac{1}{486}

= \frac{1}{486}

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What is the integral from 0 to 1/9 of x-9x^2 ?
The integral from 0 to 1/9 of x-9x^2 is 1/486