1:02 That "in terms of elementary functions" bit is like the ultimate tease. It hints that there IS a way to do it, but you'll have to delve into the abstruse and Lovecraftian world of higher pure mathematics to get your answer. The important thing here is that when you switch the order, the OUTSIDE integral must always have constant upper and lower bounds. So once you've sketched the bounds and defined the region, those bounds will drop into your lap and it's a relatively easy thing to draw the non-constant ones out of the redefined sketch. Another thing which strikes me about most of the variations on this problem you present is that the non-constant bounds of integration are usually (if not always) one-to-one functions within the region of interest and their inverses are easy to describe; I can imagine cases where things might get messy, and it's quite possible in that case that a change to polars (or from them!!!) might clear that up, but now we're getting way out of my depth and probably well beyond the bounds of the courses you're presenting here.
Marz, you're right. in the Q83 the top triangle is the one your want. The boundaries you listed on your redescribed region are also right. Thus all you need to do is reverse the order of integration, put in your new limits and integrate!
@sandslash123 If both limits were from 0 to 1 then our region of integration would be a square. See, at 3:29, the red shaded region of integration is not a square, but a triangle. How to describe the boundary of a triangle? Well, we can use the lines: y=x; y=0; x=0 and x=1. That's why functions are appearing in the limits of integration.
thank you so much! when you said that you switch which sides had the 2 parallel lines, that is when everything clicked for me. You just saved my grade on my exam which is in less then an hour :)
Great and Clear explanation, thank you so much I know understand this for my final today! I just couldn't understand it when it was explained in the book, but your presentation of the material is fantastic!
@sandslash123 Can you see that function and the region of integration are totally separate. The triangle (region of integration) has nothing to do with the function (sin x^2).
@jsm666 Yes, it can be done with triple integrals but it is slightly more challenging. I hope to post some videos about this technique for triple integrals in the future!
hey great video. I'm just having trouble interpreting the reversal of inverse trig functions. the question i have is with the bounds sin^-1(y) < x < pi - sin^-1(y), 1/root(2) < y < 1. The graph is a little trick to draw so i was wondering if there was any regorous numerical method that doesn't involve graphing? or any tricks to simplify this?
@sandslash123 "I don't see any problem integrating x with respect to x." If you mean that you don't see any problem with integrating sin(x^2) with respect to x then let me ask you this: what is the antiderivative of sin(x^2) with respect to x in terms of elementary functions?
Hey, great video. I am having some difficulty reversing the bounds of the equation with sin^-1(y) < x < pi - sin-1(y) and 1/root(2) < y < 1. I'm just having difficulty drawing the graph and interpreting the reversal of the boundaries. Is there any rigorous method that doesn't involve drawing?
Peripheral question: can the practice of changing the order of integration be extended to triple or even higher orders of integral? I imagine this would require one to be VERY careful about one's new limits.
Hello Chris I get the idea and applied it to a question I'm trying to solve, I should be getting the same value for the area of the region whether I reverse the order of integration or not, however my calculations lead me to a different value for the area of the region. Could you help me out? The region is originally defined by 0
hyeee...... rajan.....i must suggest u when u do a double triple integral ....must done frm this site...becoz in this site everything explain vry beriefly......i think if u done aw double triple frm here ,.....u must got aw gud marks in ur examss.....:)
Due to the rather simple nature of replacing x=y with y=x i am still left a bit confused on what one should do. If one had for example 0 < y < 2 (equal as well) and 0 < x < 4-y^2 would this example switch to 0 < x < 2 and 0< y < sqr(4-x).
@DrChrisTisdell Yes, ok, I seen, you are right there, but still, limits does not make any sense to me. I don't understand why are you writing some functions in the limits instead of numbers, when you can explicitly see that both limits are from 0 to 1.
I learned A level calculus, complex numbers, expansions and matrices in 3 months of uni, cuz my school didn't had these topics in syllabus, therefore I consider myself as being pretty damn blessed with math understanding, but THIS doesn't make ANY REASONABLE SENSE WHATSOEVER. This is 7th or 8th video I'm watchin'. Why the hell can't we integrate it straight away? I don't see any problem integrating x with respect to x. It is the same as say - differentiate x with respect to x - you will have 1!
I realize that this comment is several years old, but you can't find the antiderivative of sin(x^2) using elementary functions. Similar to e^(x^2). Granted, you could use Maclaurin Series for sin(x), where we'd just replace the x's with x^2's, and then integrate, but this is multivariable calculus. The point of the video is to learn how to reverse the order of integration, so that we can more easily integrate.
1:02 That "in terms of elementary functions" bit is like the ultimate tease. It hints that there IS a way to do it, but you'll have to delve into the abstruse and Lovecraftian world of higher pure mathematics to get your answer.
The important thing here is that when you switch the order, the OUTSIDE integral must always have constant upper and lower bounds. So once you've sketched the bounds and defined the region, those bounds will drop into your lap and it's a relatively easy thing to draw the non-constant ones out of the redefined sketch.
Another thing which strikes me about most of the variations on this problem you present is that the non-constant bounds of integration are usually (if not always) one-to-one functions within the region of interest and their inverses are easy to describe; I can imagine cases where things might get messy, and it's quite possible in that case that a change to polars (or from them!!!) might clear that up, but now we're getting way out of my depth and probably well beyond the bounds of the courses you're presenting here.
Wow, I went to 20 different websites and 6 other videos and yours was the only one whose explanation made sense. Thanks!
Marz, you're right. in the Q83 the top triangle is the one your want. The boundaries you listed on your redescribed region are also right. Thus all you need to do is reverse the order of integration, put in your new limits and integrate!
12 years later and this is still helpful. Thank you!
I'm glad it was useful to you. Hope you enjoy the free ebook too!
@sandslash123 If both limits were from 0 to 1 then our region of integration would be a square. See, at 3:29, the red shaded region of integration is not a square, but a triangle. How to describe the boundary of a triangle? Well, we can use the lines: y=x; y=0; x=0 and x=1. That's why functions are appearing in the limits of integration.
amazing, I love you! Can't believe I'm in my second year at lse and your teaching helped me understand this I much more!!!
Thank you soo much for this video, you break it down and show how easy it is by taking it step by step.
thank you so much! when you said that you switch which sides had the 2 parallel lines, that is when everything clicked for me. You just saved my grade on my exam which is in less then an hour :)
You're very welcome. Glad you enjoyed it.
Great and Clear explanation, thank you so much I know understand this for my final today! I just couldn't understand it when it was explained in the book, but your presentation of the material is fantastic!
My pleasure and hope you also find the ebook of some use. The free link is in the description.
You are very welcome. Lorelei.
Thanks again Dr. Tisdell! I learn more from you then my calc professor.
Happy to help! Thanks for your feedback. :-)
couldn't locate my notes for the lecture on this so it was a huge help; thanks!
@sandslash123 Can you see that function and the region of integration are totally separate. The triangle (region of integration) has nothing to do with the function (sin x^2).
@jsm666 Yes, it can be done with triple integrals but it is slightly more challenging. I hope to post some videos about this technique for triple integrals in the future!
Good to hear that this video was helpful!
hey great video. I'm just having trouble interpreting the reversal of inverse trig functions. the question i have is with the bounds
sin^-1(y) < x < pi - sin^-1(y), 1/root(2) < y < 1.
The graph is a little trick to draw so i was wondering if there was any regorous numerical method that doesn't involve graphing? or any tricks to simplify this?
@AdeliLadyOFWinds My pleasure.
You are welcome!
Glad it was of some use and hope you take a look at my new ebook also.
@sandslash123 "I don't see any problem integrating x with respect to x." If you mean that you don't see any problem with integrating sin(x^2) with respect to x then let me ask you this: what is the antiderivative of sin(x^2) with respect to x in terms of elementary functions?
This video resolved a lot of my wonders. Many thanks!!
You are very welcome!
Thank you! You made this so much easier to understand!
great video.. helped a lot.. I wish you were a maths lecturer at my uni
@cd2k85 If that was the case then you'd be integrating over a square, not a triangle. Can you see why?
Hey, great video. I am having some difficulty reversing the bounds of the equation with
sin^-1(y) < x < pi - sin-1(y) and 1/root(2) < y < 1.
I'm just having difficulty drawing the graph and interpreting the reversal of the boundaries. Is there any rigorous method that doesn't involve drawing?
You are wonderful at explaining this. Thank you so much.
Thanks. If you like this video then you also might be interested in my new ebook. It's free and the link is in the description.
Peripheral question: can the practice of changing the order of integration be extended to triple or even higher orders of integral? I imagine this would require one to be VERY careful about one's new limits.
Very helpful and informative, and well explained.
Thanks very much!!
My pleasure.
This is literally a lifesaver. Thank you!
I see now, Chris.Thanks, and keep up the good work!
You're welcome!
Hey, no worries!
Please see my other videos - I have lots including parabolas.
its really help full video
sir thank u soooooooooooooooo much for sharing
Good luck with your double integrals!
@DrChrisTisdell Ok... But, how do you know that it is a triangle? Function says that it is a sine function.
u saved me from Detention sir, i owe you :) :D
Thanks a lot! Really good explanation.
Hello Chris
I get the idea and applied it to a question I'm trying to solve, I should be getting the same value for the area of the region whether I reverse the order of integration or not, however my calculations lead me to a different value for the area of the region. Could you help me out? The region is originally defined by 0
hyeee...... rajan.....i must suggest u when u do a double triple integral ....must done frm this site...becoz in this site everything explain vry beriefly......i think if u done aw double triple frm here ,.....u must got aw gud marks in ur examss.....:)
Due to the rather simple nature of replacing x=y with y=x i am still left a bit confused on what one should do. If one had for example 0 < y < 2 (equal as well) and 0 < x < 4-y^2 would this example switch to 0 < x < 2 and 0< y < sqr(4-x).
Hi - you've got the idea in the sense that you need to invert the functions that make up the boundary of the region of integration.
life saver!!!! thanks doc
thanks so much doc, well lucid
YOU ARE AMAZING❤❤❤❤❤❤
@DrChrisTisdell Yes, ok, I seen, you are right there, but still, limits does not make any sense to me. I don't understand why are you writing some functions in the limits instead of numbers, when you can explicitly see that both limits are from 0 to 1.
thanks you so much for being so clear and awesome :D
but please tell how you integrate at last
It's OK to integrate $\sin(y^2)$ with respect to $x$, Marz, you'll get $x \sin(y^2)$, Hope that helps!
great stuff :) thanks Chris
Awesome explanation!!..:)
Hi Chris, I am not quite sure why after reversing the integral, y lies between 0 and x, why does it not lie between 0 and 1? Thanks
Oops. I didn't realize that this comment was several years old. Better late than never, I suppose.
I learned A level calculus, complex numbers, expansions and matrices in 3 months of uni, cuz my school didn't had these topics in syllabus, therefore I consider myself as being pretty damn blessed with math understanding, but THIS doesn't make ANY REASONABLE SENSE WHATSOEVER. This is 7th or 8th video I'm watchin'. Why the hell can't we integrate it straight away? I don't see any problem integrating x with respect to x. It is the same as say - differentiate x with respect to x - you will have 1!
I realize that this comment is several years old, but you can't find the antiderivative of sin(x^2) using elementary functions. Similar to e^(x^2). Granted, you could use Maclaurin Series for sin(x), where we'd just replace the x's with x^2's, and then integrate, but this is multivariable calculus. The point of the video is to learn how to reverse the order of integration, so that we can more easily integrate.
thank U very much Dr.
awesome job
Superb. Thank You.
so helpful, thank you!
@DrChrisTisdell I think that I finally got it, kind of, thx for private lesson :)
Thanks so much!
awesome sir
Brilliant!
super
thank u sir for sharing that
its really help me for understandinig..(**,)
#ftw