Tutor.com The Definite Integral Session

May. 19, 2012

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Session Transcript - Math - Calculus, 5/1/2012 10:25PM - Tutor.comSession Date: 5/1/2012 10:25PM
Length: 18.8 minute(s)
Subject: Math - Calculus


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] Find the area of the region. y = 3/x, y = 6/x^2, x = 4

Stephen L (Tutor)
[00:00:07] Hi! Welcome to Tutor.com !

Guest (Customer)
[00:00:12] hello!

Stephen L (Tutor)
[00:00:21] I see you have a problem about finding area?

Guest (Customer)
[00:00:37] yes, I have no idea how to solve this one!

Stephen L (Tutor)
[00:00:59] ok, no problem, I think I can help. Can I take a minute to work ahead though to make sure?

Guest (Customer)
[00:01:09] yes!
[00:01:35] oh!
[00:01:48] and the answer I got was 3 ln(4/3) 1/2
[00:01:52] but its incorrect

Stephen L (Tutor)
[00:02:17] ok, so you did attempt it...great!
[00:02:29] what I'm looking at right now is what the area looks like that we need to find
[00:03:15] looks to me like its a little "sliver" where the two functions cross, and bounded on its right side at x=4?
[00:03:28] I'll sketch what I think it looks like, if that's ok

Guest (Customer)
[00:03:46] okay, the graph was provided but I couldn't paste it on here

Stephen L (Tutor)
[00:04:05] ok, it will be a good check for us if I can replicate it like the graph
[00:05:43] does that look right?

Guest (Customer)
[00:05:52] yes, thats it!!!

Stephen L (Tutor)
[00:06:03] awesome!
[00:06:21] ok, and do you have a formula for calculating area?
[00:06:31] (I mean a general formula...(
[00:06:33] )
[00:07:17] it's probably something that involves the integral, right?

Guest (Customer)
[00:07:18] I do, but I used something else
[00:07:24] yes

Stephen L (Tutor)
[00:07:30] ok, what did you use?

Guest (Customer)
[00:08:28] I did that
[00:08:43] is the integral from 3 to 4
[00:08:48] or 3 to 6?

Stephen L (Tutor)
[00:08:56] nice! turns out you almost had the right formula...we just need to take a closer look at the limits
[00:08:58] :(
[00:09:00] :)
[00:09:05] sorry meant the smiley face
[00:09:20] the limits are going to be the lower and upper x-bounds

Guest (Customer)
[00:09:19] ohh okay! :)

Stephen L (Tutor)
[00:09:34] what's the upper x-bound of the area in this problem?

Guest (Customer)
[00:09:45] 4?

Stephen L (Tutor)
[00:09:50] yep!
[00:09:54] what's the lower x-bound?

Guest (Customer)
[00:10:01] 3

Stephen L (Tutor)
[00:10:11] hmmm...how did you determine that?

Guest (Customer)
[00:10:26] well thats just what I guessed

Stephen L (Tutor)
[00:11:03] oh ok. can you think of a way that we can determine the lower limit exactly?

Guest (Customer)
[00:11:08] the example has it as 2
[00:11:14] I'm not sure how to get that one

Stephen L (Tutor)
[00:11:18] yes, I think that would be right
[00:11:21] ok, no problem
[00:11:39] in this case, notice that those red and blue lines intersect on the graph
[00:11:48] the intersection is where the light green area starts
[00:12:12] to find the x-coordinate of that intersection point, we could set those two equations for y equal to each other, right?

Guest (Customer)
[00:12:34] yes!

Stephen L (Tutor)
[00:12:53] great! lets do that quickly to check that 2 is the correct value for the lower limit...can you try that?

Guest (Customer)
[00:13:24] yes,
[00:13:28] I'm doing it now

Stephen L (Tutor)
[00:13:36] ok

Guest (Customer)
[00:14:23] and I solved and got 2, and checked by plugging it back in

Stephen L (Tutor)
[00:15:18] great! nice work! I also did it and found that those functions must cross at x=0 and x=2. of course for this problem x=2 is the right one to use

Guest (Customer)
[00:15:39] yeah I was unsure about the x=0, but okay!

Stephen L (Tutor)
[00:15:46] so our limits are x=2 and x=4
[00:16:03] now your function looks good...do you understand how you got the stuff inside the integral?

Guest (Customer)
[00:16:22] yes, now we take the antiderivative and plug in the values?

Stephen L (Tutor)
[00:16:27] you got it!

Guest (Customer)
[00:16:39] I got 3ln(2)-3/2

Stephen L (Tutor)
[00:17:08] I did as well...do you have an answer to check with?

Guest (Customer)
[00:17:51] yes! I just put it into webassign and its correct!

Stephen L (Tutor)
[00:17:58] awesome!!!
[00:18:09] any questions about this? you did a great job!

Guest (Customer)
[00:18:21] thank you! no but thanks so much for the help!

Stephen L (Tutor)
[00:18:30] ok, great! have a great night!
[00:18:36] Thanks for using Tutor.com. Please fill out the survey as you leave so we can learn how we can better help you in the future! 

Guest (Customer)
[00:18:38] thanks! bye you too!

Stephen L (Tutor)
[00:18:42] bye!