# How do you find the area bounded by y^2=4x and the line y=2x-4?

Feb 10, 2018

$A r e a = 9$

#### Explanation:

Solution:

we simplify equations with respect $y$ like making $x$ output and $y$ input:
$\left({y}^{2} = 4 x\right)$$=$$\left(x = {y}^{2} / 4\right)$
and $\left(y = 2 x - 4\right)$$=$$\left(x = \frac{y}{2} + 2\right)$

we find the points of intersection of the line and parabola by solving their equations simultaneously.
${y}^{2} / 4 = \frac{y}{2} + 2$
${y}^{2} / 4 - \frac{y}{2} - 2 = 0$
$\left(y - 4\right) \left(y + 2\right) = 0$
${y}_{1} = 4$
${y}_{2} = - 2$

then we use this formula ${\int}_{c}^{d} \left[f \left(y\right) - g \left(y\right)\right] \mathrm{dy}$ to find Area between two equations

but with this formula we have some conditions
1.$f \left(y\right)$ and $g \left(y\right)$ are continuous between $d$ and $c$
2. $d \le c$
3. $f \left(y\right) \ge g \left(y\right)$ for $c \le y \le d$

then we call $d = {y}_{1}$ , $c = {y}_{2}$

and then we choose point between $d$ and $c$ to know which equation is bigger so I will choose point $y = 0$
for equation $x = {y}^{2} / 4$, $0 = {\left(0\right)}^{2} / 4$
and for equation $x = \frac{y}{2} + 2$ ,$2 = \frac{0}{2} + 2$
so ${y}^{2} / 4 \le \frac{y}{2} + 2$ for $c \le y \le d$
and then we make each equation like a function so $f \left(y\right) = \frac{y}{2} + 2$ and $g \left(x\right) = {y}^{2} / 4$
so now are ready to apply formula "${\int}_{c}^{d} \left[f \left(y\right) - g \left(y\right)\right] \mathrm{dy}$" to find Area between two equations
so ${\int}_{-} {2}^{4} \left(\frac{y}{2} + 2\right) - \left({y}^{2} / 4\right) \mathrm{dy}$ $=$ ${\int}_{-} {2}^{4} \frac{y}{2} + 2 - {y}^{2} / 4 \mathrm{dy}$
$=$ ${\left[{y}^{2} / 4 + 2 y - {y}^{3} / 12\right]}_{-} {2}^{4}$
$=$ $\left({\left(4\right)}^{2} / 4 + 2 \left(4\right) - {\left(4\right)}^{3} / 12\right)$$-$$\left({\left(- 2\right)}^{2} / 4 + 2 \left(- 2\right) - {\left(- 2\right)}^{3} / 12\right)$
$=$ $\left(\frac{20}{3}\right) + \left(\frac{7}{3}\right)$$= \frac{27}{3} = 9$