Simple mistakes | Stop THESE 3 ⚠️
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- เผยแพร่เมื่อ 28 พ.ย. 2023
- In this video, I'm going to show you three common mistakes that students make when integrating functions in calc 2. These errors can cause problems when trying to solve calculus problems, so it's important to be aware of them and avoid them! Integration can be tricky and I want to ensure you don't miss these on your next exam!
Calculus is a complex subject, and mistakes can easily lead to confusion and frustration. In this video, I'll show you how to avoid the most common integration mistakes and get the most out of your calculus studies!
Check out this playlist for more calc 2 problems!
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I learned to put the absolute symbols, but I never learned why. Thank you for the explanation.
You’re welcome
bro forgor +c
i did, y
but you asked, so i responded
Bro asked if you knew the fault drop a comment then insulted you for doing it 😂
@@misterme-sports5193 my apologies! When you started the response with bro, I thought you were trolling, hahahah! But yeah, nice job finding the error. Sorry about that
@@shadowiz6506 totally my bad. When I saw the "bro" in the response, I assumed he was trolling and not answering a question. My mistake
@ the original poster -- The host of this site is not your "bro."
In fact, the shear symbol "Integral(1/x, dx)" is not well defined if you take seriously a definition of the indefinite integral. One essential factor is missed in most materials/books when dealing with it. Namely, to have a nice theory you start with a function f:D -> R which must be defined on a connected set D, which in the case of reals means an interval. Then, the proposition which claims that any two antiderivatives are equal up to some constant C is true. So in this perspective when we write Integral(1/x, dx) in fact it is a shorthand for two integrals: Int(f1(x), dx) with f1:{x >0}->R, where f1(x)=1/x or Int(f2(x), dx) wher f2(x)=1/x. Remember 1/x as R\{0}->R is not defined on a connected set, thus we take as a domain either {x > 0} or {x < 0}.
Well done. Yeah, we can think of it as an adjoined piecewise function
Actually, the indefinite integral of 1/x is:
log|x| + C(x)
where the log is natural and C is a locally constant function (hence C(x)= a constant for x0).
You are correct! Great explanation
Great explanation
Glad it was helpful!
4:43 that kind of mistake can happen with derivatives too.
technically hte two pieces of ln|X| can also have different constants. So it's really f(x) = ln(x) + c if x > 0 and ln(-x) + d if x < 0.
Yes, absolute value function can be represented by a piecewise function
@@NumberNinjaDave I think that you're missing the main part of their comment. The absolute value doesn't change the value of the constant. For example, f(x) = ln(x) + 6 if x > 0 and ln(-x) + 3 are is also an antiderivative for 1/x, even when the constants are different.
@@boston4635 you are correct though you can indeed use a piecewise function to help represent this visually. And the anti derivative is a family of functions so any constant is just a member in the family of solutions
ln |x| has a different meaning than ln x with x being restricted to x > 0. One accepts all x except 0, and turns negative x into positive, the other is not valid for x
That’s correct, although it’s good to understand why the integral gives the absolute value, which is the point of the video. If we blindly just memorize the integration, we will miss out on the domain concept
@@NumberNinjaDave If you use complex log instead of just the real numbered natural log, you have a function to use that doesn't require the extra absolute value signs. Complex log(z) = ln(|z|) + i*(angle(z) + 2*pi*k). Letting k=0, and using C - pi*i as the constant of integration, we can connect this to the standard form used in Calc 1.
@@carultch sure, though in this video, I'm only focused on R as the domain
OK. But instead of adding the constant can we subtract a constant?
Lol
@@NumberNinjaDave I hope you can laugh more, here:
Place an infinite amount of points on a circumference of a circle. Then pick any point of your choice on the circumference. Add one to that point or subtract one from that point. How far have you moved on the circumference in radians?
@@user-ky5dy5hl4d sounds like you already know the answer 😊
@@NumberNinjaDave I love physics. But unfortunately one cannot do anything on physics without mathematics. But I am not a mathematician by any means. I just dabble in math. My answer is such that if you add one on that circumeference you will not move at all because infinity is not a number. But one more question (from physics). What causes the speed of light?
@@NumberNinjaDave Sir! I thank you for your input. I am learning from you. I had had heavy calculus years ago. But now people like you retreive my mind about math. I believe that you are an ingredient to have an element about time which can be expressed in math. You are great!
2:55 ->
+ cockroach?
Lizard actually 🦎
@@NumberNinjaDave 😂😂😂😂
@@notdeus3834 I think lizard math would be tricky. When the lizard sheds its tail, it's now a piecewise function
forgot +c
update: WOOOOOOOO 🤪🔥💯💪
haha
Nice work!
👍
👍
A common mistake even among mathematicians, is not realising the constant can change beyond discontinuities.
For example, the proper antiderivative of 1/x is ln(x) + c1 when x>0, and ln(-x) + c2 when x
Another way of thinking about it, is that the “c” is just a function that has a derivative of 0 along all the places where f(x) is defined. But it can still suddenly change values where f(x) and F(x) is undefined.
That’s a piecewise definition that is essentially how the domain of the absolute value in ln | x| + C is defined
Very true!
Who put tĥat poor bird over there ?
Toucan Sam 🦅