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Change the order of integration to solve tricky integrals
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- เผยแพร่เมื่อ 17 ส.ค. 2024
- The double integral over a region can be expressed in two different ways. It could be that we write dxdy which means that we integrate with respect to x first and then y (called horizontal strip). Or it could be dydx which means y is integrated first and then x (vertical strips). The big idea is we can convert between these and so can choose whichever is the easier of the two. The steps are:
1) Given an integral, sketch the region implied by the limits of integration
2) Now interpret that region the opposite way, such as vertical strips instead of horizontal strips
3) Write the new integral in the opposite order of integration
4) Hopefully it will now be easier to compute!
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I wasn't understanding how to assign the values of limits and I was worried about it as my exams are approaching. Now I have understand it clearly Thank you Dr. Bazett, you are a life saver. Love from India.
me too. actually i need more videos on this topic but don't know what to search
Yes, Indian math professors focus more on solving the questions rather explaining what's happening.
IITB? EP?
This man will single handed save my calculus journey.
I’m so glad I stumbled upon this channel, it’s a gold mine. Great work!
This video gives me hope. I was on the verge of breaking into molecules as I did not understand a single word from my Instructor when he explained this. Thank you very much, and now I get it.
You explained this so clearly. I’m so happy I found this channel for calc 3
Man you opened my eyes and i finally can understand the logic behind this. That's very great teaching
I am serious. Thank you, i am an ee student and my professor just taught us how to do this very fast like everyone would know. you're a king, God bless u
Sometimes maths really is like magic! Such a wonderful technique. Thanks for the video. Pure class.
Superb explanation and very succinct too. Great video Dr. Bazett.
Great video and explanation!! Thank you!!
Glad it helped
Something needed now that everything is online due to corona until next year
Omg this saved my life!!!! thankyou so much, i had so much trouble visualizing the inner and outer parts on a diagram and now i understand
Thank you! Please keep uploading such wonderful videos!
great video. All other instruction videos did not go as deep to the basis as this one, I can say for sure.
can you explain why for dx we use horizontal strips ? My confusion stems from the fact that in single variable calculus when we integrate f(x)dx we view the function as being divided into many vertical strips. I am confused as why in this double integral we use horizontal strips for dx and in single variable integral we used vertical strips.
You may consider the horizontal strip dy being partitioned into several blocks along x-axis. Integrating x first means that these blocks is sumed into the strip dy according to the basic concept of integration. Integrating y then sum up the strip dy into the area contained in the region.
Hi Dr Trefor Bazett. Thanks for the great explanation and solid visuals!
I'm a little confused on one part and was hoping you could help clear up my misunderstanding 😁
Why is the upper bound on the inner x integral y^3 and not 8?
We have the line x=8 plotted and it's equivalent to 2^3.
Why can we not argue that the largest x value of the shaded area is 8 instead of y^3?
Because then you would be integrating the box enclosing the region instead of the region itself. The switched integral is saying "for each y from 0 to 2, add up all the cross-sections of the function at that y where x is between 0 and y^3." Each 2D slice has a different width at each y value. There is one cross-section where the width is 8, and that's when y = 2. But the cross-section at y = 0 has a width (x) of 0. So for y = 0, you have to add 0 area instead of a cross-section 8 wide.
why did you not change the bounds to match with u = y^4+1?
Really it's a masterpiece!! ❤
love you boss!
this is just what i needed
Very thorough tutorial!!! Enjoyed watching!
The hardest part, I've found, is to know how to plot these functions in order to see where is the region you want to integrate. I need to build geometric intuition.
Thank you. This concept is simple yet hard to visualize sometimes
It's not obvious, but the factorization of y^4+1 over the reals is (y^2−y√2+1)(y^2+y√2+1); it's the difference of the two squares (y^2+1)^2 and 2y^2.
Of course, if you then use partial-fraction decomposition and complete the square and find an antiderivative, you'll end up with a sum of logs and arctans that would be unusually difficult to work with, and *then* you would need to deal with the compositions of those functions with the cube root to calculate the outer integral.
To spell it out more, it turns out that an antiderivative of 1/(y^4+1) is sqrt(2)/8*(ln(y^2+sqrt(2)y+1)-ln(y^2-sqrt(2)y+1)+2arctan(sqrt(2)y+1)+2arctan(sqrt(2)y-1)); evaluated at the upper limit, this is sqrt(2)/8*(2pi-2arctan(2sqrt(2)/3)+ln((33+20sqrt(2))/17)), and it turns out that when you evaluate it at the lower limit, the resulting function of x *does* have an elementary antiderivative, but it's far from obvious, and it also is far from obvious that every term cancels except for a couple of terms that add up to ¼ln(17).
You're an excellent teacher.
After integrating the inside integral, which is a constant 1/y^4+1 how did you plug in y^3 on the top? It is a constant number (1) on the top, why is that one being replaced with the bounds of integration? (y^3) Doesnt it need to be a y in order to be replaced with y^3?
Dr. Trefor the GOAT 🐐
Sir Excellent video!
Keep doing such videos 🔥
You are a mathematics maniac
Thanks
you are just awesome and genius.
sorry but i don't understand why integrating with respect to y first would imply the vertical strips and not the horizontal ones...i mean if you see it as as sums of little rectangles (riemannian integral i guess) the dy should imply small rectangles with infinitesimal height and width equal to the value of the function in x...I'm just a bit confused.Thanks for the video though..very informative
These integrals are always with respect to two variables, just a question of order. I.e dydx or dxdy. Whn doing dydx this is saying first dy then dx. So when doing the first dy you are changing the y values i.e. it is a vertical strip. That strip yes will have a width of dx, but we are imaginging changing it's long height into little segments of height dy and integrating to get the full vertical strip. Then integrated w.r.t dx to add up all the vertical strips.
@@DrTrefor Thanks for the reply...I think i got ..still need to wrap my head around it for a while but i guess i can manage that :D
Fantastic video, thank you.
You've done your bit to make life on this planet a little less miserable, thank you
Could you make one for triple integrals please! Great vids
Nice simple and clear explanation, thanks for the help!
Thank you sir 🙏🙏🙏
Thank you again man, you are saving my career! :D
th-cam.com/video/IangXACFW48/w-d-xo.html
why do you not replace the upper and lower bounds for dy when doing u-substitution? where if you plug in you would get y =1 and y = 16?
Yoo, I appreciate this!
how to proceed if we are having 2 regions and which to select
Awesome! Thank you.
Please more vidéos of exercices like this one
Thank you so much sir
Really helpful for calculus 1 thanks
Who would dislike lmao. He makes the topic so simple.
I remember my first hater who went through every single video I had (back when they had like less than a hundred views) and dislike every single one in about 10 minutes:D AFter that I really don't care anymore lol
@@DrTrefor it's probably him😂
Thank you sir!
I have a question.
Let's say I wanna solve
Integral from 0 to 1 of the integral from 0 to √(x²+y²) of 1/y⁵+1 dy dx . How can I solve this integral? If I wanna change the order, but I have a two variable function in the inside integral. So can I use polar coordinates in this integral? It looks it's gonna go bad.
helpful 🙏
Is there any analytical algorithm to do this ? Because sometimes bounds may be complicated functions to draw !!
Absolute legend :D
Great professor
an actual g
i never comment bruh but ur a life saver holy shit
This was really nice.
Great work Dr
Dude you are fucking awesome! I finally get it now! I hope I get at least a B in the class so I can stay in the honors program at my university!
why didn't you change the interval after substitution?
very helpful
Good explanation
Ty!
Thanks a Lot Proffesor. im from Srilanka
god tier explanation
th-cam.com/video/IangXACFW48/w-d-xo.html
MUCH THANKS
Doesn't integrating with respect to y, mean that you have horizontal strips? since dy is the width of the strips this means that the strip is horizontal (given that dy is y2-y1)
Thanks again
this guys cadence is hard
thanks a lott sir for such wonderful vids
Doctor, what if the root of x is a substitution of the cube root of x, what would the solution be?
my question is how am i supposed to know what the graph looks like on an exam (especially sqrt(x). is there a mathematical way to figure the new bounds for integration??
Legend 😊😊💝💝💖💖
💝Sir I am big fan of you💝
Could you please explain changing of triple integral as well
Nyc explanation... But try to stretch the length of the video by increasing quantity of videos as well as Difficulty level ! Thnks.
How would you calculate the center of a force field, is there any method?
Nice explanation
Very nice video!
Hmm, I don't get what this integral represents... There is no x in the function, so what volume is this?
sometimes the function is really weird and I can't draw it, any tips?
Well done 👍
you are awesome
Hi Trefor. Do you intend to make videos on the vector calculus section of Calculus III and the V.C. integral theorems? Thanks!
@@DrTrefor Excellent! Thank you in advance.
Speak is most powerful
Great!!!!!
AMEN FOR YOUR VIDEO I GOT STUCK ON A COSX/X when we didnt even discuss yet Taylor series. finally got a way to solve my problem by doing the changing of order. All this through your help. Tysm!
Nice video 👍👍
Int(1/(1+y^4))=Int(y^-2/(y^-2 + y^2))=1/2Int((1+y^-2)/((y-y^-1)^2 +2)-1/2Int((1-y^-2)/((y+y^-1)^2 +2))
Some nice subs appear, which eventually lead to a solution...
Yeah maybe switching order first is nicer lmao
Great
It is possible to solve the integral of 1/x⁴+1 dx!
😊😊
Can you remake a video about strong induction? The video has low quality. Anyway, I liked the video.
Trefor Bazett Thank you. Keep doing what you are doing.👍
❤
😭🙏
JAI SHREE RAM
Man integration is too hard
يا شين هذرتك اللي م لها داعي
Almost great, just slow down a bit.
Just do the problem! Too much talking
Thanks
❤