# The curve y=3x-x^2 cuts the x axis at the points O and A and meets the line y=-3x at the point B, as in the diagram. a) calculate the coordinate of A and B? B)FIND THE AREA OF SHADED REGION. i got the coordinates but the area is the issue

## Jun 12, 2018

$\textcolor{b l u e}{A = {\int}_{0}^{6} \left(\left(3 x - {x}^{2}\right) - \left(- 3 x\right)\right) \cdot \mathrm{dx} = 108 - 72 = 36}$

#### Explanation:

a)
to calculate the point A ,O and B

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

${x}^{2} - 6 x = 0 \Rightarrow \left({x}^{2} - 6 x + 9\right) - 9 = 0$

${\left(x - 3\right)}^{2} = 9 \Rightarrow \left(x - 3\right) = 3$

"when" $x = 6 \Rightarrow y = - 18$

$B = \left(6 , - 18\right)$

$O = \left(0 , 0\right) \mathmr{and} A = \left(3 , 0\right)$

b)
Now lets calculate the area of the shaded region.

The Area due to x-axis given by:

color(red)[A=int_a^b(y_2-y_1)*dx

$A = {\int}_{0}^{6} \left(\left(3 x - {x}^{2}\right) - \left(- 3 x\right)\right) \cdot \mathrm{dx}$

$A = {\int}_{0}^{6} \left(6 x - {x}^{2}\right) \cdot \mathrm{dx} = {\left[3 {x}^{2} - \frac{1}{3} \cdot {x}^{3}\right]}_{0}^{6}$

$= 108 - 72 = 36$ 