Use y as the variable of integration to find the area between the graph of y = 1/x and the y axis from y = 1 to y = infinity
Ok. y = 1/x so x = 1/y. The length of each "rectangle" should be x and the width should be dy.
So it should be the limit of ∫1/y dy from y = 1 to y = b as b approaches infinity, = lim (ln b - ln 1 ) as b approaches infinity, = ∞
Rewrite the integral found in the preceding problem using x as the variable of integration.
Ok first I tried using the change of variable system, then rewriting from scratch. Both produced the same answer. Rewriting from scratch:
The area should be ∫ x dy from y = 1 to y = infinity. dy = -1/x^2 dx so we can replace this with:
∫ -1/x dx from y = 1 to y = infintiy
Lets reverse the evaluation limits to make it positive:
∫ 1/x dx from y = infinity to y = 1
Ok. From y = infinity to y = 1. y = 1/x so x = 1/y when y = infinity, x aproaches zero to the right.
When y = 1, x = 1 so we should be able to rewrite this as:
limit of ∫ 1/x dx from x = b to x = 1 as b approaches zero from the right. (This should be the answer to the problem.)
= limit of [ ln 1 - ln b ] as b approaches zero from the right.
= 0 - (-∞) = ∞
In both cases, we ended up evaluating ln ∞ + ln 1. Well technically we ended up with - ln 1 in the first, but thats the same as ln 1/1 = ln 1 = 0. So it ended up being the same thing.
The bizzare answer my book gave is as follows:
The limit of ∫ (1/x - 1) dx from x = 0 to x = 1 as b approaches zero from the right.
Whhhaaaat??? As b approaches zero from the right? I don't see a b in that expression at all! And where did that -1 come from?
Either its a misprint or I am doing something COMPLETELY wrong.
Last edited by mikau (2006-02-25 06:56:51)