Question
Find the exact area between y=lnx and y=ln(x^ 2 ) for 4≤x≤8 .
Answers
Steve
just plug and chug. The area is
∫[4,8] ln(x^2)-lnx dx = ∫[4,8] 2lnx-lnx dx = ∫[4,8] lnx dx
Now, using integration by parts,
u = lnx, du = 1/x dx
dv = dx, v = x
∫lnx dx = x lnx - ∫ dx = x lnx - x
So, the area is
(8ln8-8)-(4ln4-4)
= 24ln2 - 8 - 8ln2 + 4
= 16ln2 - 4
∫[4,8] ln(x^2)-lnx dx = ∫[4,8] 2lnx-lnx dx = ∫[4,8] lnx dx
Now, using integration by parts,
u = lnx, du = 1/x dx
dv = dx, v = x
∫lnx dx = x lnx - ∫ dx = x lnx - x
So, the area is
(8ln8-8)-(4ln4-4)
= 24ln2 - 8 - 8ln2 + 4
= 16ln2 - 4