you will need the intersection of
y = lnx and y = 2-lnx
2lnx= 2
lnx = 1
x = e
so you will have a region shaped like an "arrow head" from x=e to 4
Area = ∫(2 - 2lnx) dx from x = e to 4
(remember the integral of lnx is
xlnx - x , which should be in your repertoire of common integrals )
= [-2xlnx + 4x] from e to 4
= -8ln4 + 16 - (-2e + 4e)
= .....
Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to four decimal places.)
Enclosed by y = ln x, y = 2 − ln x, and x = 4
1 answer