use geometry not integration by parts!!

Рет қаралды 66,186

Michael Penn

Күн бұрын

Пікірлер: 139

@GroundThing 4 ай бұрын

>"Integral of ln(x) without integration by parts" >*Looks inside * >Integration by Parts

@ElusiveEel 4 ай бұрын

there's one without integration by parts, where you use the chain rule on xln(x) hold on, chain rule is also integration by parts

@tinkeringtim7999 Ай бұрын

> integration by parts > *_looks inside_* > geometry

@idolgin776 4 ай бұрын

I like it. Very log-ical.

@anis786 4 ай бұрын

And natural. ln-ical

@jewulo 4 ай бұрын

😁

@tommasotiberi5666 3 ай бұрын

This is going in my best comments cata-log

@Player_is_I 3 ай бұрын

This comment is so punny

@mekaindo 3 ай бұрын

@@Player_is_I colorful reply huehuehuehue

@Vega1447 4 ай бұрын

Very neat. The trick of "turning the xy axis sideways" always appeals.

@brettaspivey 4 ай бұрын

That is what integration by parts is!

@Rockyzach88 4 ай бұрын

They never explain why it's called that (the derivation I supposed) in undergrad math. Or at least I don't remember doing that.

@BrianGriffin83 4 ай бұрын

He says that at 7:24

@BrianGriffin83 4 ай бұрын

Also, this is a particular case of IBP (i.e. when one of the "factors" is the identity function).

@ib9rt 2 ай бұрын

The method is called "integration by parts" because it involves dividing the rectangle into the sum of two parts! This is probably the best way to introduce integration by parts to students. It makes it much easier to comprehend.

@glenm99 3 ай бұрын

This is how I have thought of integration by parts since the day I first saw it. uv = integral of fdu + integral of gdv, where f and g are the same curve on the plane, just viewed from a different axis. If the function isn't invertible but is still "nice," you can break it into parts and add them up, or else apply a transformation and later its inverse. It all washes out in the end. But I've asked around and never heard of it taught this way.

@henrycook859 4 ай бұрын

Huh, I felt like I could understand this geometric argument even with very little background of calculus. Thanks!

@loopingdope 3 ай бұрын

Nice

@element1192 3 ай бұрын

I love derivations like this that take the logical path rather than the short path

@titan1235813 4 ай бұрын

The e^y integral is the genius part. That's the key there!

@MarcoMate87 4 ай бұрын

Fantastic approach. Indeed, this technique can be applied whenever the integrand is invertible.

@jackkalver4644 4 ай бұрын

This is a special case of integration by parts that’s easy to prove: inverse functions. Can you use geometry to show that if y=f(x) then int y dx=xy-int x dy?

@MarcoMate87 4 ай бұрын

I think that what you wrote is correct, using precise hyphotesis of course, and the proof is identical to that shown in this video.

@kinexkid 4 ай бұрын

Its also so beautifully elegant when you can take an expression that most people would just chug along with the basic way to evaluate it, and chamge it into something with geometry. To me, geometry is the most elegant way to translate pure math into something visual for even a layman to understand

@goodplacetostop2973 4 ай бұрын

7:41

@benedictrodil4931 4 ай бұрын

Always the comedian

@aayush554 4 ай бұрын

never change

@rcallenberg 4 ай бұрын

The shown calculation is for x >=1. In order to be complete - wouldn't it be necessary to look at the integral between 0 and 1 as well?

@nadavslotky 3 ай бұрын

That is left as an exercise for the avid viewer 😉

@Nameless-qe9hu 3 ай бұрын

I think the process is the same, just replace “1” with an arbitrary “c” value

@aNu-9017 4 ай бұрын

Understood almost the entire calculus class with that video!

@212ntruesdale 4 ай бұрын

Extremely clever, thoroughly enjoyed. Thanks for sharing!

@FineFlu 3 ай бұрын

this is so much more intuitive and tangible than the typical presentation

@Pictoo-ID 3 ай бұрын

That's exactly how I integrate ln(x) for the first time in my life. It makes me cry of happiness when I watch you doing the exact solution I did in the past. Amazing❤

@bubbotube 4 ай бұрын

1:34 I'm actually a proponent of the name 'natural exponential' for the function exp.

@txikitofandango 4 ай бұрын

As soon as you drew the picture, there it was

@Rockyzach88 4 ай бұрын

Cool shit. I have a math minor and learned something. I always want to go back through calculus completely just because I know I could probably understand more things and get a deeper understanding overall.

@sonicmaths8285 2 ай бұрын

Get a good book abt calc for example from Spivak and do the exercises. The exercises are crazy but great to become much better at it

@kingplunger1 4 ай бұрын

Very cool. I like it when you can use geometry instead of the usual appraoch, especially when it turns out to be relatively simple

@laitinlok1 3 ай бұрын

I mean it's basically calculating the area between t=x and t =e^y , so it is basically integrating (x-e^y) between y=0 anf y= ln x which is equal to (xy -e^y) when y is between 0 and ln x. Substituting it would show (xlnx - e^ln x) - [x(0) - e^0]. Simplifying it, it shows xln x - x + 1.

@DrR0BERT 4 ай бұрын

I have added your video to my on-line class videos as we head into Integration by Parts.

@Inspirator_AG112 3 ай бұрын

*[**0:15**]:* Not if you have the 3rd dimension! (:

@utuberaj60 4 ай бұрын

Nice one Michael. In fact integration by parts is intimidating,to say the least, for anyone. Another method I saw on Blackpen Red Pen, but of course with limits of 0 and 1 for the integration of ln x, the area under this region is easily evaluated using the idea that e^x and ln x are inverses of each other,and we look at rhe graph of e^x and evaluate it for the said limits and get thr answer, without resorting to IBP. Anyway, your method is more specific to the indefinite integral of ln x.

@BrianGriffin83 4 ай бұрын

More generally, this is a nice method to find an expression for the indefinite integral of the inverse of a bijection f, although it's a bit cumbersome: calling g the inverse of f and F an antiderivative of f, it's xg(x)-F(g(x))+C.

@capnbug 3 ай бұрын

I agree, brian

@BikeArea 4 ай бұрын

So elegant and quite intuitive! 👌✌️

@RiteshPhysics 3 ай бұрын

Same can be done with sin^-1

@tpthpt5973 3 ай бұрын

Great video!

@priyanshsrivastava3571 2 ай бұрын

u just used used the property of int(a to b) f(x) dx + int(f(a) to f(b)) f inverse (x) dx = b*f(b)-a*f(a) , but with whole proof and a geometrical feel ... THAT EXPLANATION WAS A NICE ONE !!

@rick4135 3 ай бұрын

Please introduce riemman stiltjies integral and some examples!!!! Great video

@scottmiller2591 3 ай бұрын

Yes, this is exactly how I think of integration by parts, as you say in your summary.

@r.i.p.volodya 3 ай бұрын

Clever - I'd not seen that before.

@Jack_Callcott_AU 4 ай бұрын

Molto semplice, molto carino !

@jaimeduncan6167 4 ай бұрын

I see no problem finding the derivative of ln(x) outside an exam, it's pretty obvious almost everybody will try with x*ln(x) see that you get an extra one, and subtract an x. The first time I encountered it was because of the rocket equation, and wanting an equation for the distance under constant gravitation during my high school years. Once you get it you derive and get it. Clearly in the University, they will not accept that it's obvious and you will need to use integration by parts. This is clever, on the other hand, and a beautiful showing of trying a different approach that could be useful in other cases when the antiderivative is not obvious.

@sayantanroy-o4s 3 ай бұрын

You can put lnx=z and x=e^z lnxdx becomes e^zdz , then no integration by parts is required

@BenyKarachun 3 ай бұрын

No, it becomes the integral of z*(e^z)dz, and then you still have to do integration by parts

@sayantanroy-o4s 3 ай бұрын

@@BenyKarachun no , you can make it (z+1-1)e^z then substitute ze**z =p

@martinr7728 3 ай бұрын

Am I missing something? What about the area from 0 to 1?

@luismuller6505 4 ай бұрын

Thats funny for me to see you make a video on this because I actually came up with basically this exact argument for the antiderivative of y = ln(x) a cuple of months ago when I was still in 12'th grade.

@OssamaAzuzaui 3 ай бұрын

What if 0

@ElusiveEel 4 ай бұрын

I stumbled upon this other derivation when doing exercises on differential equations y = x*ln(x) dy/dx = ln(x) + 1 integrate both sides with respect to x y = ∫ln(x)dx + x + c ∫ln(x)dx = y - x + c ∫ln(x)dx = x*ln(x) - x + c

@pouet4608 4 ай бұрын

great one. thanks.

@jmafoko 3 ай бұрын

amazing stuff

@benardolivier6624 4 ай бұрын

I remember this proof from somewhere else, and it was shown by a 12 year old kid. (maybe from BPRP channel)

@robertpearce8394 4 ай бұрын

Maybe from the paper that Michael Penn clearly referenced.

@user-SK22-calc 4 ай бұрын

Hey micheal. Will you maybe in the future have a linear algebra playlist on the second channel?

@UltraMaXAtAXX 4 ай бұрын

He already does, in a way.

@user-SK22-calc 4 ай бұрын

@@UltraMaXAtAXX ?

@eeerrrrzzz 4 ай бұрын

How convert the final result to indefinitive integral I don't get it.

@hrperformance 3 ай бұрын

Very nice

@sergiogiudici6976 4 ай бұрын

How could It be extended to much more inverse Function integral?

@gregwochlik9233 4 ай бұрын

Nice one!

@moonwatcher2001 3 ай бұрын

Awesome ❤

@vdinh143 3 ай бұрын

While this is very ingenious, it's not any easier than IbP in the sense that it's not any more likely for an average student to figure out on their own; in fact it could be said to beway harder considering how none of us have seen it before. However, this technique can be generalized into every InP problem so it is definitely a better way to teach students to solve this class of problems OVER memorizing uv - vdu

@diondrefuentes4384 2 ай бұрын

Does this work for trig functions

@michel_dutch 4 ай бұрын

Very nice!

@General12th 4 ай бұрын

Very cool!

@gabrielbarrantes6946 2 ай бұрын

This can be generalized to any monotonous function.

@reapicus557 4 ай бұрын

Fantastic video! I will always appreciate a new addition to my toolset. Plus, for some reason, I always do the integral of ln(x) in my head when I need a distraction, and this way is much more fun that by parts. ദ്ദി ˉ͈̀꒳ˉ͈́ )✧

@BeatsNotBears 4 ай бұрын

Can someone explain where we hid the integration by parts? Was it in chamging the variable of finding the area of region 2?

@woody442 4 ай бұрын

This approach doesn't need integration by parts (which is the 'classic' method of solving this). That was the point of the video to offer a solution without it.

@DeanCalhoun 4 ай бұрын

Essentially, the xlnx term on the rhs is the uv part of the integration by parts formula and the area of R2 is the integral of vdu term which has been moved to the other side and had a substitution applied (which is equivalent to the geometric argument of taking the integral along the y axis). The integration by parts formula comes directly from integrating the product rule for derivatives: (uv)’ = u’v + uv’. Again, this matches up exactly with Michael wrote after you consider the substitution which has been couched in a geometric argument.

@ProactiveYellow 4 ай бұрын

This geometric argument is where integration by parts comes from. For parameters u and v, we can break up the rectangle uv into two parts, one measured with respect to the v axis (the integral of u dv) and one measured with respect to the u axis (the integral of v du). We can write this as follows: uv=int(u dv)+int(v du). The exact divide of the region depends on what u and v are, but in this example, you get a natural log splitting the region. The traditional method is u=ln(x) and v=x, but really it's u=ln(x), v=e^(ln(x)). You'll find that u and v are commonly expressible in a composition like this so you're relating the parameters.

@BeatsNotBears 4 ай бұрын

@@ProactiveYellow Ahh you're right,, I was having trouble seeing udv and vdu. Thanks!

@woody442 4 ай бұрын

@BeatsNotBears I entirely miscalculated the level of complexity you expected from an answer. Sorry for that, and thanks for all the great explanations of how this is actually related to the 'classic' method.

@AhmadJamal-v4y 3 ай бұрын

Ok that is kinda cool

@ArdiSatriawan 2 ай бұрын

That is a geometric interpretation of integration by parts 😂

@archangecamilien1879 3 ай бұрын

Let me guess, lol...use the symmetry of ln(x) and e^x, etc...I mean, they are symmetric about the origin or something, etc...pretend one is really integrating e^x with the necessary modifications...

@johnchristian5027 3 ай бұрын

beautiful way to do it without integration by parts

@pabloduchen3842 2 ай бұрын

He did use integration by parts. The video title is misleading and clickbait.

@pocojoyo 3 ай бұрын

Suggestion: ser x1.25 speed

@genres381 4 ай бұрын

use geometry to find integral of 1/x

@MyeonggyuRyu 4 ай бұрын

Brilliant

@pankajk.r2448 3 ай бұрын

Nice 👍👍

@Calcprof 4 ай бұрын

Integration by parts is often badly taught and motivated. Take a look at ow physicists typically use it --often not to integrate specific functions, but to manipulate various integrals to get insight. Integral f g' = - integral f'g + boundary terms. Easy!

@LoneDevelopper 3 ай бұрын

Actually i thought of this years ago, damn

@ominollo 4 ай бұрын

Nice 👍

@cameronspalding9792 4 ай бұрын

If I wasn’t allowed to use integration by parts: my first choice would be substition

@montasir0607 3 ай бұрын

Good luck doing $x^2sinx dx

@barryzeeberg3672 4 ай бұрын

can this technique be applied to other functions? if not, what is the speciaproperty of ln that makes it feasible here?

@GroundThing 4 ай бұрын

In fact it can. Say you want to take the integral of a generic function f(t). The rectangle would be xf(x), region 1 would be your desired integral and region 2 would be the integral from 0 to f(x) of f^-1(t)dt. Performing a u substitution, for f(u)=t you get dt=df(u), so the integral of udf(u) from 0 to x. Putting that all together you get your desired integral as x*f(x)-integral of xdf(x), renaming the dummy variables. Now let's do a change of variables, for x=v and f(x)=u, and all together you have integral of udv= uv - integral of vdu. This is just integration by parts.

@barryzeeberg3672 4 ай бұрын

@@GroundThing Thanks, that is interesting. I was wondering, the integration of region 2 (ie, using the original y axis as if it is the x axis for the integration step) reminds me a little of Lebesgue integration? Is this correct, or am I off base?

@gabrielbarrantes6946 2 ай бұрын

Calculus II? In community college maybe lol

@arthursteenkist2595 4 ай бұрын

Cool!!

@AnatoArchives 4 ай бұрын

base

@Happy_Abe 4 ай бұрын

Doesn’t this geometric setup assume x>1? What happens when x

@chemicalbrother5743 4 ай бұрын

That is the part under the x-axis and left from the function, so its not part of the rectangle and we can ignore it.

@bjornfeuerbacher5514 4 ай бұрын

@@chemicalbrother5743 That is only not part of the rectangle for x > 1. For x < 1, the rectangle will lie below the x-axis.

@bjornfeuerbacher5514 4 ай бұрын

@Happy_Abe: Yes, this assumes x > 1. For x < 1, you'll have a rectangle below the x-axis and can use a similar argument there.

@allanjmcpherson 4 ай бұрын

It still works, but it gets a little messy. You have two regions between the curve and the x-axis. One region has signed area x ln x < 0, and the other region has signed less than zero that can be computed as minus the integral of (e^y -1) from 0 to ln x. We need to add in the minus sign since we the signed area is negative with respect to the x-axis, but we're integrating with respect to y, and the curve is "above" the y-axis. Once you account for this, you end up with x ln x - x + C as expected.

@Happy_Abe 4 ай бұрын

@@bjornfeuerbacher5514 thanks that’s what I was looking for?

@hskstudies2412 3 ай бұрын

Happy teachers day 😊

@martinkunev9911 3 ай бұрын

this doesn't explain the indefinite integral

@cjvoids2994 3 ай бұрын

I’d rather by parts

@TheBluePhoenix008 3 ай бұрын

This feels like it's been made for children. What level is this class?

@Will-nf9gf 4 ай бұрын

d/dx( f(x) * ln(x) + g(x) ) = lnx, f(x)/x + ln(x)f'(x) + g'(x) = ln(x), let f(x) = x, f(x)/x = 1, ln(x)f'(x) = ln(x), so 1 + ln(x) + g'(x) = ln(x), g'(x) = -1, g(x) = -x + C, so f(x) * ln(x) + g(x) = xln(x) - x + C ... this is a weird solution i came up with that sort of feels wrong. under what conditions can you write a function as the derivative of the product of itself and another function plus another function?

@IoT_ 4 ай бұрын

Just try to search for the proof of the theorem of integration by parts. It's proven exactly using the product rule for the differentiation

@Will-nf9gf 4 ай бұрын

@@IoT_ then maybe i found a weird version of integration by parts lol but ill try to stick to the general formula

@carultch 4 ай бұрын

This follows a similar idea as a way to derive integration by parts in general, from geometric reasoning: kzbin.info/www/bejne/fIfIe5mXatCIh6c

@jmafoko 3 ай бұрын

cant integration by parts problem be transformed into similar form, even better give a proof integration by part using elementary geometry

@xizar0rg 4 ай бұрын

Is the title of the video correct? They're normally useless but this is particularly poorly formed. (Title at time of posting is "use geometry not integration by parts!!"