How do you find the area between #g(x)=4/(2-x), y=4, x=0#?

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Answer 1 · 2016-11-15T11:02:08

The area is #=4-4ln2#

A small area is given by #dA=(4-4/(2-x))dx#

The point of intersection of #g(x)# and #y=4#

is #4=4/(2-x)# #=>#, #2-x=1# #=>##x=1#

Therefore, the area is

#A=int_0^1(4-4/(2-x))dx=[4x+4ln(2-x)]_0^1 #

#=4+0-0-4ln2=4-4ln2#

graph{(y-4/(2-x))(y-4)=0 [-7.1, 6.95, -0.03, 6.99]}