"The integral of x^(-1) seems to have followed the regular rule, but is in fact an exception..." -- It is such gems of your comments which make your videos incredibly useful and a treat to watch!
It's been 5 years or more since I last dealt with Calc. I just started working on my BSEE and they started throwing calc based formulae at us in a 101 class and for the life of me I couldn't even remember where to start. I've spent my weekend watching this series up to ep. 20 now and taking notes and working through all of the practice problems just like if it were a real class, and I can't thank you enough for the way you present the information. Though I have learned all this some years ago, I don't remember ever understanding it so well. Calc was always a steep uphill struggle of thinking I understood a thing only to find I'd done everything fundamentally wrong, repeat. So far, following through your videos, I haven't gotten a single comprehension check wrong, which is extremely satisfying. Thank you, and please do continue.
I thought i was about to fail the FRQ portion on the Calc test tomorrow, I didn't know about this channel's existence, you literally explained a topic i didn't understand as good as I do now.
If you expand the domain to complex numbers, you'll see precisely why the integral of 1/x dx is ln|x| + C, and why it still works for -x values. In general, a complex logarithm, ln(z) is equal to ln(mod(z)) + i*arg(z)*k. The function mod(z) is the modulus of z, which is an extension of absolute value, for real numbers, where it tells us how far from the origin the number is, as a positive number. The function arg(z) is the argument of z, which is the angle CCW from +real, that locates the complex number, and k is any integer. When z is a negative real number, arg(z) = pi, and the function reduces to ln(|z|) + i*pi*k. The i*pi*k can therefore be part of your +C, and the log of a negative number has a real part that starts with the log of its absolute value. The other part of your +C, can nullify this imaginary part, if you strategically select the right constant. It is a half truth to say that the integral of 1/x dx is ln|x| + C, but it keeps it simple. In reality, the +C can be different on both sides of zero, but it usually doesn't make a difference when calculus is limited to real numbers.
My humble request to professor dave that if he slowly understand his video to everyone than it will be more helpfull.I have not understand because your video & the subtitle are so first for me to grapple with. Because i am a beginner.. Who have no idea about integration. 😭. Plz understand it more easily so that everyone can get it & you are a very good teacher.. 😊
@@Zone_Ranger You also cannot integrate 1/x across the singularity of x=0. When evaluating ln(abs(x)) to determine the integral of 1/x, both limits of integration have to be on the same side of zero.
You're using ambiguous notation: at 6:01, 2 2/3 isn't the same as 2 * 2/3. The first evaluates to 8/3, the second evaluates to 4/3. I recommend to always use a multiplication sign if needed.
@@ProfessorDaveExplains Your explanation doesn't contain a scientific proof for implying multiplication if any two terms are adjacent. What's the purpose of mixed numbers? For showing the whole part and the fraction. It's more complicated to see the whole part in an improper fraction. Putting myself into your shoes, I could tell that nobody uses cosecans and secans past like America although it makes easier to write trigonometric expressions.
You think I'm bragging that I'm not in fourth grade? How about this, find a mixed number in any calculus textbook anywhere, take a picture of it and email it to me. If you can do that, I'll apologize profusely for this exchange.
Nice... Actually Integral of X^n = (X^n+1) /n+1 +C even for n=-1 I(X^n) = (X^n+1) /n+1 +C just that C = -1/(1+n) +C2 I =Ilm n-> -1 [ (X^n+1) -1]/n+1 = exp [lnX*(n+1)]/(n+1) = In(x)
Outa curiosity for the prob at 8:50 could that “1/3” been brought back up to the variable x So itd look more like: ln(x)^1/3 Or i guess now that i think about it…. 1/3lnx could ideally be written as: ln cuberoot of x ?? I imagine Im not to sure
To determine the indefinite integral of ln(x) dx, you use integration by parts. Assign: u = ln(x) dv = dx Therefore: du = 1/x dx v = x integral u dv = u*v - integral v du ln(x) * x - integral x * 1/x dx ln(x) * x - integral 1 dx Result: ln(x) * x - x + C
Taking online classes for engineering calc and the teacher in the video for this part is just awful. He babbles, mumbles and stutters. If he ever went over this at all I totally missed it - even though it's a video and it was a question on the test, this helped me get it. So thanks.
@@abhilashasinha5186 The short answer is yes. The Calc 1/Calc 2 level answer is: most of the time, yes. The Calc 3 and DiffEQ answer is, definite integrals can also be functions, and very often are. If the function doesn't have any variables in it, other than the variable of integration, then a definite integral is always a number, in contrast from an indefinite integral being a family of functions that only differ by the arbitrary constant. There are applications of definite integrals, where the function being integrated either contains a constant unrelated to the variable of integration, or one or both of the limits of integration are functions, such as multi-dimensional integrals. There are also applications in vector calculus, where the arbitrary constant of integration, is actually an arbitrary function of integration.
What he means, is there is no product rule that always gets you the answer. Integration by parts works to undo the product rule of differentiation, but it isn't a method that always works, even if you can integrate both component functions. It works in special cases, where it simplifies the functions to a form that can ultimately be integrated. Either the ender, the looper, or the regrouper.
Professor Dave over here saving my Calc grade.
"The integral of x^(-1) seems to have followed the regular rule, but is in fact an exception..." -- It is such gems of your comments which make your videos incredibly useful and a treat to watch!
indeed!
It's been 5 years or more since I last dealt with Calc. I just started working on my BSEE and they started throwing calc based formulae at us in a 101 class and for the life of me I couldn't even remember where to start. I've spent my weekend watching this series up to ep. 20 now and taking notes and working through all of the practice problems just like if it were a real class, and I can't thank you enough for the way you present the information. Though I have learned all this some years ago, I don't remember ever understanding it so well. Calc was always a steep uphill struggle of thinking I understood a thing only to find I'd done everything fundamentally wrong, repeat. So far, following through your videos, I haven't gotten a single comprehension check wrong, which is extremely satisfying. Thank you, and please do continue.
binge watching your calculus playlist and your videos made me understand calc a lot better than my professor did for months. thank you so much!
You are not only the master at math (and various other subjects), but you are also the master at teaching and making people understand.
someone once told him to enunciate and boy, did he take it to heart!
I thought i was about to fail the FRQ portion on the Calc test tomorrow, I didn't know about this channel's existence, you literally explained a topic i didn't understand as good as I do now.
I still don't understand why you have negative votes. It's beyond my comprehension. I love your videos!
Dave...you are amazing, I it seems like a chunk when learning but so much easier when you explain!!!
I watched it two times and I took notes.
This is often the missing part of my calculus! Thank you professor for explaining me this important and often missed idea!
I really like how you explain things
Thank you sir for your dedication and for making this free! 🙏
Thanks Dave for the quick rundown. I'm taking an at home quiz and needed a refresher.
Try (x^h-1)/h on a graphing calculator. For small h it looks more and more like ln x
in the second example, 1/x can be a negative number, but in the antiderivative ln x, x must be positive, so they are not equivalent.
If you expand the domain to complex numbers, you'll see precisely why the integral of 1/x dx is ln|x| + C, and why it still works for -x values.
In general, a complex logarithm, ln(z) is equal to ln(mod(z)) + i*arg(z)*k. The function mod(z) is the modulus of z, which is an extension of absolute value, for real numbers, where it tells us how far from the origin the number is, as a positive number. The function arg(z) is the argument of z, which is the angle CCW from +real, that locates the complex number, and k is any integer.
When z is a negative real number, arg(z) = pi, and the function reduces to ln(|z|) + i*pi*k. The i*pi*k can therefore be part of your +C, and the log of a negative number has a real part that starts with the log of its absolute value. The other part of your +C, can nullify this imaginary part, if you strategically select the right constant.
It is a half truth to say that the integral of 1/x dx is ln|x| + C, but it keeps it simple. In reality, the +C can be different on both sides of zero, but it usually doesn't make a difference when calculus is limited to real numbers.
12 year old obsessed (kind of) about maths is here to thank you for this wonderfull serie!
I wish I was obsessed with math in your age. I would have it so easier going through my Master’s exam…
Incredible videos. It would be pretty cool if you were to do a video on differential equations.
Hes covered differentiation before integration???
My humble request to professor dave that if he slowly understand his video to everyone than it will be more helpfull.I have not understand because your video & the subtitle are so first for me to grapple with. Because i am a beginner.. Who have no idea about integration. 😭. Plz understand it more easily so that everyone can get it & you are a very good teacher.. 😊
I am at a point where i dont even know how to thank you....superb staff!!!!
You the real MVP.
the integral of 1/x is ln|x|+c. (you need to include the absolute value )
in fact if I wanted to be more precise we would have that the integral of 1/x is ln(x)+c1 for x>0 and ln(-x)+c2 for x
whoops! i know i included that in another tutorial, must have slipped my mind for this one.
@@Zone_Ranger You also cannot integrate 1/x across the singularity of x=0. When evaluating ln(abs(x)) to determine the integral of 1/x, both limits of integration have to be on the same side of zero.
You're using ambiguous notation: at 6:01, 2 2/3 isn't the same as 2 * 2/3. The first evaluates to 8/3, the second evaluates to 4/3. I recommend to always use a multiplication sign if needed.
Nobody uses mixed numbers past like 4th grade. Multiplication is implied when any two terms are adjacent.
@@ProfessorDaveExplains Your explanation doesn't contain a scientific proof for implying multiplication if any two terms are adjacent. What's the purpose of mixed numbers? For showing the whole part and the fraction. It's more complicated to see the whole part in an improper fraction. Putting myself into your shoes, I could tell that nobody uses cosecans and secans past like America although it makes easier to write trigonometric expressions.
Scientific proof? Um, this is math, bud. I'm just telling you that nobody uses mixed numbers in real math. Do with that information what you will.
@@ProfessorDaveExplains What is real math according to your definition? I use mixed numbers sometimes. Am I nobody? Or are you just bragging?
You think I'm bragging that I'm not in fourth grade? How about this, find a mixed number in any calculus textbook anywhere, take a picture of it and email it to me. If you can do that, I'll apologize profusely for this exchange.
I'm just imagining e^x, on that ladder you showed on the previous episode. And thinking, "You fool! It's e^xs all the way down!"
Until d/dy shows up.
Hey dave, can you do a evaluation of the definite integral by limit of riemann sums?
I did that earlier in the series! Check out "what is integration".
If online school was more like Prof Dave's explanations, maybe it wouldn't be so bad.
Fact
very nice explanation
u the best teacher ever!! ❤️🙏🙌🤸
thank you
Tnx professor Dave 👍👍👍
This is nice Mr Dave
Thanks Prof dave
You are a life saver
Nice...
Actually Integral of X^n = (X^n+1) /n+1 +C even for n=-1
I(X^n) = (X^n+1) /n+1 +C
just that C = -1/(1+n) +C2
I =Ilm n-> -1 [ (X^n+1) -1]/n+1 = exp [lnX*(n+1)]/(n+1) = In(x)
f'(x)= nx^n-1
thank you so much!!
Done.
Outa curiosity for the prob at 8:50 could that “1/3” been brought back up to the variable x
So itd look more like:
ln(x)^1/3
Or i guess now that i think about it….
1/3lnx could ideally be written as:
ln cuberoot of x
?? I imagine
Im not to sure
your making it way more difficult than is should be.
It doesn't get any simpler than this.
Bro tell me how I come here to understand the work but end up looking at his merchandise instead.
actually too relatable
Nice sir
Guys, what's the antiderivative of In x dx.
x ln(x) - x + C
To determine the indefinite integral of ln(x) dx, you use integration by parts.
Assign:
u = ln(x)
dv = dx
Therefore:
du = 1/x dx
v = x
integral u dv = u*v - integral v du
ln(x) * x - integral x * 1/x dx
ln(x) * x - integral 1 dx
Result:
ln(x) * x - x + C
Taking online classes for engineering calc and the teacher in the video for this part is just awful. He babbles, mumbles and stutters. If he ever went over this at all I totally missed it - even though it's a video and it was a question on the test, this helped me get it. So thanks.
so indefinite integrals of functions are always functions and definite integrals of a function are always numbers?
CALCULUS
thanks issac newton
for making me a 12th grade mathematician
@@abhilashasinha5186 The short answer is yes. The Calc 1/Calc 2 level answer is: most of the time, yes. The Calc 3 and DiffEQ answer is, definite integrals can also be functions, and very often are.
If the function doesn't have any variables in it, other than the variable of integration, then a definite integral is always a number, in contrast from an indefinite integral being a family of functions that only differ by the arbitrary constant.
There are applications of definite integrals, where the function being integrated either contains a constant unrelated to the variable of integration, or one or both of the limits of integration are functions, such as multi-dimensional integrals. There are also applications in vector calculus, where the arbitrary constant of integration, is actually an arbitrary function of integration.
Nice
5:09 What do you mean there is no product rule for integration?! What is integration by parts then?
What he means, is there is no product rule that always gets you the answer. Integration by parts works to undo the product rule of differentiation, but it isn't a method that always works, even if you can integrate both component functions. It works in special cases, where it simplifies the functions to a form that can ultimately be integrated. Either the ender, the looper, or the regrouper.
so...can you be my math teacher lol
Love you STEM Jesus
What if You have dX in the equation
dx is always in the equation, it represents the end of the integral and it is necessary for integration.
Being the 2K(th) person liking this video, I feel powers within
Sorry fast
wow，you speak like a native speaker!
The 11 people who disliked are mad he explained it better than their teachers.
After watching this as an Indian students.....i just wish I could study in American schools....i could literally tore apart the exams😂.....
Great tutorial