Asked by Anonymous
How do you find the area enclosed by y=lnx , y=1-lnx, and x=4.
Answers
Answered by
oobleck
The region has vertices at (√e,1/2), (e,0), (4,ln4), (4,0)
Since the boundary is not a single function, the area will be the sum of two regions.
If you use vertical strips, the boundary changes at x=e
A = ∫[√e,e] (lnx-(1-lnx)) dx + ∫[e,4] lnx dx
= 2√e - e + ln256 - 4 = 2.124
Using horizontal strips,
A = ∫[0,1/2] 4-e^(1-y) dy + ∫[1/2,ln4] 4-e^y dy
= 2+√e-e + ln256+√e-6 = 2.124
Since the boundary is not a single function, the area will be the sum of two regions.
If you use vertical strips, the boundary changes at x=e
A = ∫[√e,e] (lnx-(1-lnx)) dx + ∫[e,4] lnx dx
= 2√e - e + ln256 - 4 = 2.124
Using horizontal strips,
A = ∫[0,1/2] 4-e^(1-y) dy + ∫[1/2,ln4] 4-e^y dy
= 2+√e-e + ln256+√e-6 = 2.124