Question

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

Answer 1

`Area =9`

Explanation:

Solution:

we simplify equations with respect `y` like making `x` output and `y` input:
`(y^2=4x)` `=` `(x=y^2/4)`
and `(y=2x-4)` `=` `(x=y/2+2)`

we find the points of intersection of the line and parabola by solving their equations simultaneously.
`y^2/4=y/2+2`
`y^2/4-y/2-2=0`
`(y-4)(y+2)=0`
`y_1=4`
`y_2=-2`

then we use this formula `int_c^d[f(y)-g(y)]dy` to find Area between two equations

but with this formula we have some conditions
1.`f(y)` and `g(y)` are continuous between `d` and `c`
2. `d<=c`
3. `f(y)>=g(y)` for `c<=y<=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=(0)^2/4`
and for equation `x=y/2+2` ,`2=(0)/2+2`
so `y^2/4<=y/2+2` for `c<=y<=d`
and then we make each equation like a function so `f(y)=y/2+2` and `g(x)=y^2/4`
so now are ready to apply formula "`int_c^d[f(y)-g(y)]dy`" to find Area between two equations
so `int_-2^4(y/2+2)-(y^2/4)dy` `=` `int_-2^4y/2+2-y^2/4dy`
`=` `[y^2/4+2y-y^3/12]_-2^4`
`=` `((4)^2/4+2(4)-(4)^3/12)` `-` `((-2)^2/4+2(-2)-(-2)^3/12)`
`=` `(20/3)+(7/3)` `=27/3=9`