Added: 3 years ago
From: DrChrisTisdell
Views: 13,508
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  • Great explanation. I've ploughed through a number of videos on this topic and your video was the only one that finally made things click.

  • Brilliant!

  • Thanks a lot! Really good explanation.

  • @AdeliLadyOFWinds My pleasure.

  • 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 are wonderful at explaining this. Thank you so much.

  • Thanks again Dr. Tisdell! I learn more from you then my calc professor.

  • 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.

  • @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!

  • u saved me from Detention sir, i owe you :) :D

  • I see now, Chris.Thanks, and keep up the good work!

  • 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

  • @cd2k85 If that was the case then you'd be integrating over a square, not a triangle. Can you see why?

  • 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!

  • @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?

  • @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.

  • @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.

  • @DrChrisTisdell Ok... But, how do you know that it is a triangle? Function says that it is a sine function.

  • @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).

  • @DrChrisTisdell I think that I finally got it, kind of, thx for private lesson :)

  • This is amazing. Thank you!

  • Thank you! You made this so much easier to understand!

  • Life-saving!

  • Comment removed

  • Great video mate..simple and easy to understand!

  • Awesome video, really helped me out. Thanks!

  • Thanks!  That was really clear and easy to follow!

  • #ftw

  • Nicely done Dr. Tisdell, congratulations on this clear explanation,

    Jose - University of Wisconsin-Madison

  • Hey Chris, thanks so much for your help on this!!!

  • thank u sir for sharing that

    its really help me for understandinig..(**,)

  • its really help full video

    sir thank u soooooooooooooooo much for sharing

  • Wow, you're SO much better than my lecturer. Our lecturer just assumes everyone's as smart as he is :(

  • Thank you so much for this video !

  • 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?

  • 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?

  • U made it so simple.. Thanks!

  • You are welcome!

  • hey chriss

    im having problems with reversing the order of intergration

    in question 83 in the question booklet, i find after i have done all my reversing that i still have to intergrat sin(y^2) which is not cool.

    the boundries im getting are 0<=y<=12 and 0<=x<=y/3 which i think is wrong, but cant figure out what it whould be.

    from my diamgram i get a triangle and my shaded in regoin is the top one

    am i doing this correctly cuz its doing my head in?

    thank you.

    btw this video did help alot!

  • It's OK to integrate $\sin(y^2)$ with respect to $x$, Marz, you'll get $x \sin(y^2)$, Hope that helps!

  • 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!

  • couldn't locate my notes for the lecture on this so it was a huge help; thanks!

  • You're very welcome. Glad you enjoyed it.

  • awesome! this video was so much help!

  • Hey, no worries!

  • Wow, I went to 20 different websites and 6 other videos and yours was the only one whose explanation made sense. Thanks!

  • Happy to help! Thanks for your feedback. :-)

  • Thank you for this, it really helped.

  • Good to hear that this video was helpful!

  • Great Help Thanks, DR.Chris!!!!!!!!!!!

  • You're welcome!

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