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

Question

Aurora Alvarado · Accepted Answer

#Area =9#

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#

Elijah Abram · Answer

undefined To find the area bounded by $ y^2 = 4x $ and the line $ y = 2x - 4 $, you first need to determine the points of intersection between the two curves by setting them equal to each other and solving for $ x $. Then, you integrate the difference between the two functions with respect to $ x $ within the bounds of intersection points. This integral represents the area enclosed between the curves.