What is the integral from 0 to C of (1-r/C)r ?

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

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integral from 0 to C of (1-r/C)r

\int_{0}^{C} (1 - \frac{r}{C}) r d r

\frac{C ^{2}}{6}

Solution steps

\int_{0}^{C} (1 - \frac{r}{C}) r d r

Apply Integration By Parts

: [\frac{r ^{2} ( C - r )}{2 C} - \int - \frac{r ^{2}}{2 C} d r]_{0}^{C}

= [\frac{r ^{2} ( C - r )}{2 C} - \int - \frac{r ^{2}}{2 C} d r]_{0}^{C}

\int - \frac{r ^{2}}{2 C} d r = - \frac{r ^{3}}{6 C}

= [\frac{r ^{2} ( C - r )}{2 C} - (- \frac{r ^{3}}{6 C})]_{0}^{C}

Simplify

[\frac{r ^{2} ( C - r )}{2 C} - (- \frac{r ^{3}}{6 C})]_{0}^{C} : [\frac{3 C r ^{2} - 2 r ^{3}}{6 C}]_{0}^{C}

= [\frac{3 C r ^{2} - 2 r ^{3}}{6 C}]_{0}^{C}

Compute the boundaries:

\frac{C ^{2}}{6}

= \frac{C ^{2}}{6}

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