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#