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.
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.
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?
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
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.
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.
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❤
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.
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.
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.
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 !!
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.
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
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.
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.
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 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.
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
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!
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.
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...
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.
@@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?
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 )✧
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?
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!!"
![Как изобретают/открывают математику? [3Blue1Brown]](http://i.ytimg.com/vi/M0eSkRjn7h8/mqdefault.jpg)
>"Integral of ln(x) without integration by parts"
>*Looks inside *
>Integration by Parts
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
> integration by parts
> *_looks inside_*
> geometry
I like it. Very log-ical.
And natural. ln-ical
😁
This is going in my best comments cata-log
This comment is so punny
@@Player_is_I colorful reply huehuehuehue
Very neat. The trick of "turning the xy axis sideways" always appeals.
That is what integration by parts is!
They never explain why it's called that (the derivation I supposed) in undergrad math. Or at least I don't remember doing that.
He says that at 7:24
Also, this is a particular case of IBP (i.e. when one of the "factors" is the identity function).
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.
Huh, I felt like I could understand this geometric argument even with very little background of calculus. Thanks!
Nice
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.
I love derivations like this that take the logical path rather than the short path
The e^y integral is the genius part. That's the key there!
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?
I think that what you wrote is correct, using precise hyphotesis of course, and the proof is identical to that shown in this video.
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
Fantastic approach. Indeed, this technique can be applied whenever the integrand is invertible.
7:41
Always the comedian
never change
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?
That is left as an exercise for the avid viewer 😉
I think the process is the same, just replace “1” with an arbitrary “c” value
Understood almost the entire calculus class with that video!
As soon as you drew the picture, there it was
this is so much more intuitive and tangible than the typical presentation
*[**0:15**]:* Not if you have the 3rd dimension! (:
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.
Extremely clever, thoroughly enjoyed. Thanks for sharing!
Very cool. I like it when you can use geometry instead of the usual appraoch, especially when it turns out to be relatively simple
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.
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
I have added your video to my on-line class videos as we head into Integration by Parts.
1:34 I'm actually a proponent of the name 'natural exponential' for the function exp.
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❤
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.
I agree, brian
Great video!
Yes, this is exactly how I think of integration by parts, as you say in your summary.
So elegant and quite intuitive! 👌✌️
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.
You can put lnx=z and x=e^z lnxdx becomes e^zdz , then no integration by parts is required
No, it becomes the integral of
z*(e^z)dz, and then you still have to do integration by parts
@@BenyKarachun no , you can make it (z+1-1)e^z then substitute ze**z =p
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.
Same can be done with sin^-1
Clever - I'd not seen that before.
Please introduce riemman stiltjies integral and some examples!!!!
Great video
What if 0
Molto semplice, molto carino !
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 !!
Am I missing something? What about the area from 0 to 1?
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.
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
amazing stuff
great one. thanks.
How could It be extended to much more inverse Function integral?
I remember this proof from somewhere else, and it was shown by a 12 year old kid.
(maybe from BPRP channel)
Maybe from the paper that Michael Penn clearly referenced.
Does this work for trig functions
How convert the final result to indefinitive integral I don't get it.
That is a geometric interpretation of integration by parts 😂
Nice one!
Good video
Very nice!
Very nice
Can someone explain where we hid the integration by parts? Was it in chamging the variable of finding the area of region 2?
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.
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.
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.
@@ProactiveYellow Ahh you're right,, I was having trouble seeing udv and vdu. Thanks!
@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.
Hey micheal. Will you maybe in the future have a linear algebra playlist on the second channel?
He already does, in a way.
@@UltraMaXAtAXX ?
beautiful way to do it without integration by parts
He did use integration by parts. The video title is misleading and clickbait.
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
Very cool!
Ok that is kinda cool
This can be generalized to any monotonous function.
If I wasn’t allowed to use integration by parts: my first choice would be substition
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!
use geometry to find integral of 1/x
Actually i thought of this years ago, damn
Awesome ❤
Doesn’t this geometric setup assume x>1?
What happens when x
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.
@@chemicalbrother5743 That is only not part of the rectangle for x > 1. For x < 1, the rectangle will lie below the x-axis.
@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.
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.
@@bjornfeuerbacher5514 thanks that’s what I was looking for?
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...
can this technique be applied to other functions? if not, what is the speciaproperty of ln that makes it feasible here?
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.
@@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?
This feels like it's been made for children. What level is this class?
Calculus II? In community college maybe lol
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 )✧
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?
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
@@IoT_ then maybe i found a weird version of integration by parts lol but ill try to stick to the general formula
Brilliant
base
Suggestion: ser x1.25 speed
this doesn't explain the indefinite integral
Nice 👍👍
I’d rather by parts
Nice 👍
Good luck doing $x^2sinx dx
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!!"
Happy teachers day 😊
Maybe I have a deja-vu, but think this was in the channel before...
This follows a similar idea as a way to derive integration by parts in general, from geometric reasoning:
ruclips.net/video/FVeJfa5kUQE/видео.html
so any integral of f(x) = x*f(x) - f-1(x)+C ?🤣🤣🤣🤣🤣
Cool!!
cant integration by parts problem be transformed into similar form, even better give a proof integration by part using elementary geometry
Actually, when replacing ln x with an arbitrary invertible function, this is essentially the geometric interpretation of integration by parts. Nice.
Nah; use monte carlo.
Literally uses harder integration methods for no god damn reason lmao
What you did is literally the definition of integration by parts. Your video title and thumbnail are just clickbait.
52 sec ago posted
Please write ln(x) not ln x. Makes it hard to read
Pls pin