How do you find the area between #y=x, y=2-x, y=0#?

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Answer 1 · 2017-06-11T18:31:08

Let #A# be the area bounded by #y=x,y=2-x,# and #y=0#

#=>#
#A=int_0^1int_y^(2-y)dxdy=1#

First, a good thing to do would be sketch the graph.

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So, we are looking for the area of that triangle.

Take note of the points of intersection.
Here we are dealing with #y=2-x# and #y=x#
So we set both equations equal and get

#x=2-x#

#<=># add #x# to both sides

#2x=2#

#<=># divide both sides by #2#

#x=1#

So, both functions intersect at #(1,1)#

Since we are only dealing with values greater than #0# we will have some #int_0^1fdy#

And since our area is bounded on the left by #y=x# and on the right by #y=2-x#, If we let #A# be our area we get

#A=int_0^1int_y^(2-y)dxdy#
#A=int_0^1int_y^(2-y)dxdy=int_0^1(2-y-y)dy=int2-2ydy#
#=2int1-ydy=2(y-y^2/2)]_0^1=2[(1-1/2)-(0-0/2)]=2[1/2]=1#