# Calculate the area between the curve y=3x-x^2 and the x axis between x=10 and x= 6?

Jan 13, 2018

$A = \frac{493}{3}$

#### Explanation:

$y = 3 x - {x}^{2}$

The graph of this function looks like this.

graph{3x-x^2 [-3.46, 11, -100, 10]}

All of the area between $x = 6$ and $x = 10$ is below the axis so we do need to correct for this. The area is given by:

${\int}_{6}^{10} 3 x - {x}^{2} \mathrm{dx} = {\left[\frac{3}{2} {x}^{2} - \frac{1}{3} {x}^{3}\right]}_{6}^{10}$

Now evaluate the limits:

$\left\{\frac{3}{2} {\left(10\right)}^{2} - \frac{1}{3} {\left(10\right)}^{3}\right\} - \left\{\frac{3}{2} {\left(6\right)}^{2} - \frac{1}{3} {\left(6\right)}^{3}\right\}$

$= \frac{3}{2} \left(100\right) - \frac{1}{3} \left(1000\right) - \frac{3}{2} 36 + \frac{1}{3} 216$

$= - \frac{496}{3}$

It is negative because the area is under the curve, we can simply to take the positive to get:

$A = \frac{493}{3}$