How to change the order of a double integral 

DOUBLE INTEGRAL A BEGINNING FINALLY

Will you ever upload the full version of the outro song? Also please make more differential equations, they were really awesome and i'm kind of struggling with the subject right now, anyway, great videos, you make the most scary questions so understandable is amazing

Nice example about how to define a region of integration. The explanation was a bit confusing though. I would have drawn the triangle right from the beginning, explain the original limits of integration by "painting" the triangle with horizontal lines, and then introduce the new limits of integration with vertical lines. Anyway, you ended up using the series, so I can't see the benefit of changing the order of integration...

Calc 3 let's goooo

I have 2*ln(2)-1 for the 2nd integral

That problem from the coment can be easily done using the taylor expansion for ln(1+x) and than divide that by x and integrate from 0 to 1 and you get 1/1^2-1/2^2+1/3^2-... And we know that 1/1+1/2^2+1/3^2+.... Is pi^2/6 and from that you deduce that that series which is exactly like sum of squares inverse but it's alternating with +- is pi^2/6(half of the original series)

i watched 5 different videos i only understood with your video

Mr blackpenredpen, I`ve problem with understanding nyquist stability criterium. Would you mind showing, how to create nyquist plot and properly understand what is shown there. I don`t really get it.

Can u do some basic multivariable videos for students who finished calc 2

blackpenbluepen

I figured the integration limits without a picture

Do you know for what N is Ln(N) transcendental?

Am I mistaken? But why not use U-sub for ln(x+1)/x

11:15 wondering!

1:02 I actually want to have dy first. Y? Because *I WANT* to have.

Is this an April fools joke!

Why i cant understand that i try but I always fail🥲

Can you do a triple integral

How is the integral from 0 to 1 of ln(1+x) equal to ln(4)-1 and not ln(2) ? Am I missing something ?

I solved it without changing the order. The inside integral is just a function of y, let's call it f. We can integrate f by parts, just like lnx, to get yf(y) minus the integral of yf'(y)=ln(1+y) From 0 to 1 we get f(1) minus the integral of ln(1+y) where f(1) is just the integral of ln(1+x)/x and the rest is the same.

Not all heroes wear capes. Do you teach at a university? If you do I want to travel the world and study at your uni.

I can't stop watching ur vids. They're awsome. Thanks, and greetings from a spanish mathematician

The video is 11:11 :O Make a wish!

Excellent videos 3blue one brown is also good channel. I still find it hard to understand what you are saying sometimes. Could you explain in general what double and triple integral means geometrically?

Wouldn't you mind expplain rules when i can swap Integration order and rules when i can swap integration and summation?

Thank you so much 🙏

Ln(1+x)/x can be integrated using parts, or DI method. Think of it as x^-1(ln[1+x] . Now we diffrentiate the log part using the DI method, and integrate the x^-1 . This works out nicely as easier perhaps than power series...

you're a genius

Bravo!!

Is it possible to figure out the bounds of integration without having to graph the region of integration?

why the limit of y cannot go from o to x when we swapped to dydx ? please tell as i took it from 0 to x and i am getting only ln4-1 as final answer.

at thge end you asked if we need +C. The answer is absolutely NO!. This is a proper integral so we are evkuating the area under the curve. The +C would cancel out, so to put +C is wrong.

Nice, close to 100k subscribers! Early congratulations. Love this channel

Damn, so many of these are underviewed

One of my favorite videos! Thanks!

Really showing integrity.

Great video

when doing it by myself, i started with differentiation under the integral sign where I(a)=$(0,1,($(0,y,ln(ax+1)/x)dx)dy). then, skipping a bit you can do yourself, i found I(1)=$(0,1,[(1+a)ln|1+a|/a]-1)da-. from there, I used the same method as previously, and stated F(b)=$(0,1,[(1+a)ln|1+ab|/a]-1)da. solving that, (subbing u for 1+ab stumped me for a bit,) i ended up getting F'(b)=1, F(1)=%(0,1,1)db-%(0,1,1)da=0. if i got 0 as my final answer, can anyone follow my process and tell me where I went wrong? I know i didn't give specifics, but between what steps did i make my mistake? id go into detail but it takes up a page and a half.

how do evaluate integral 5t^2-3t+18/t(9-t^2) shoes limits 2 goes to 1 please jaldi solution btain

can you find the double integral of modulus (xy)dxdy over region ellipse and circle.

iterated integration?? o.o

Is this a Papa Flammy reference?

Great problem! Being 10 years removed from Calculus BC I had forgotten the power series for ln(1 + x)...

Please can you solve this integration :- root Sin(x) dx ??

Cool jacket where id u get it? :)

wow

Great!

nice, thank you so much!

This proof is absolutely beautiful!

Very well explained, always enjoy your videos

Can I give you a list of hard limits to do without l'hopital unless you need to? I recommend you to get B.Demidovitch problems book

The answer to the first integral was already given as a condition for evaluating the double integral, but glad you worked out the series part.

Isnt the area of the icosoles right triangle one half?

Where is a link to the video about power series for ln?

black pen blue pen :-3

10:21 YOU used "is nothing but" like Flammable maths does! What does that phrase even mean? People are nothing but atoms but saying "integral is nothing but..." Is like saying "I am nothing but hungry" which doesn't make much sense

Upper bounds are invisible in integral !!! I didn't understand the integral in first, therefore Video has wrong !

Hey blackpenredpen, Nice Video! I've recently got a problem on an integration problem and I cannot figure it out. I would like to ask if you could try this problem. :( Evaluate the integral from 0 to infinity of (exp(3x) - exp(x))/[x(exp(3x) +1)(exp(x)+1)] Thank you

I have little question when I calculate the integrate ln(1+x) I used the U substitution to let U=1+X to transfer the question became ln(u) then integration by part but when I put the boundary in the answer why my answer is different with yours?? Then I use integration by part directly let u=ln(1+X) du=1/(1+X) so I can get the right answer finally was that wrong if I use integration by part so I can not use U substitution first?? Or do i miss to consider other condition?