# How do you find the area between f(y)=y(2-y), g(y)=-y?

Jun 3, 2018

$\frac{9}{2} u n i t {s}^{2}$

#### Explanation:

Claculating the intersection Points:
$x \left(2 - x\right) = - x$
so we get
$x \left(3 - x\right) = 0$
So we have
${x}_{1} = 0$
or
${x}_{2} = 3$
and we have
${\int}_{0}^{3} \left(x \left(2 - x\right) + x\right) \mathrm{dx} = {\int}_{0}^{3} \left(3 x - {x}^{2}\right) \mathrm{dx} = \frac{9}{2}$

Jun 3, 2018

$A = {\int}_{0}^{3} \left(3 y - {y}^{2}\right) \cdot \mathrm{dy} = {\left[\frac{3}{2} {y}^{2} - \frac{1}{3} {y}^{3}\right]}_{0}^{3} = \frac{9}{2}$

#### Explanation:

The area between two curves due to y-axis is given by :

color(red)[A=int_a^bx_2-x_1*dy

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

${x}_{1} = - y$

lets find the cross between the curves:

$2 y - {y}^{2} = - y \Rightarrow {y}^{2} - 3 y = 0 \Rightarrow y \left(y - 3\right) = 0$

$y = 3 \mathmr{and} y = 0$

$A = {\int}_{0}^{3} \left(2 y - {y}^{2}\right) - \left(- y\right) \cdot \mathrm{dy}$

$A = {\int}_{0}^{3} \left(3 y - {y}^{2}\right) \cdot \mathrm{dy} = {\left[\frac{3}{2} {y}^{2} - \frac{1}{3} {y}^{3}\right]}_{0}^{3} = \frac{9}{2}$

show the wanted area below (shaded):

${x}_{2} = 2 y - {y}^{2}$ red curve(green)

${x}_{1} = - y$ blue curve