Simple mistakes | Stop THESE 3 ⚠️
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- Опубликовано: 28 сен 2024
- 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
The absolute value symbols are what Calc books and Calc teachers tell you to do to keep it simple, but it still isn't complete.
You can have a different constant of integration on both sides of the singularity. So it really should be a piecewise function with two constants of integration, instead of a one-size-fits-all.
If you go beyond the real numbers, the integral of 1/z dz is complex log(z) + C. The log of z is ln|z| + 2*pi*k*angle of Z, where k is any integer. Evaluating across any two values of z is path-dependent, as it depends on how many complete cycles you make, to get between the lower bound and upper bound.
So we see that if we set the constant for the negative numbers equal to C - pi*i, and set the constant for the positive numbers equal to C, that we get back the familiar version, ln|x| + C.
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
technically using the complex definition of ln(x) the domain is expanded to include the negative numbers and as far as real numbers the imaginary parts which will come with this definition of ln(x) will cancel out in any definite integral as it will always be pi*i, hence in this context where you are allowing logs of negative numbers shouldn't ln(x) be fine without the modulus?
@@abcalphabet this video is focused on R, not the generalized a+bi abscissa and ordinate representation of a number where a non zero b would have a complex plane projection. The default modus operandi ( pun intended) for this course curriculum, unless otherwise noted, focuses on R.
But, that could be a topic for another future video idea!
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
Great explanation
Glad it was helpful!
forgot +c
update: WOOOOOOOO 🤪🔥💯💪
haha
Nice work!
Its ln|x| + c right?
@@abhipatel4595 yup!
@@NumberNinjaDave ty!
This is misinformation. The natural logarithm can take negative inputs. The outputs would just be complex numbers, which is handled by the "constant of integration.". The constant of integration is also not a true constant, as the function 1/x is undefined at x=0. Hence, the constant of integration should be a locally constant function which is constant on (-infinity,0) and on (0,infinity). The values of this locally constant function on these two intervals are not necessarily equal.
@@Snookimichev it’s not misinformation if my video by convention restricts the domain to the set of real numbers, just like we conventionally see in classrooms. What you’re describing is more advanced, which I don’t cover in my video. For the set R, this is the correct way.
Who put tĥat poor bird over there ?
Toucan Sam 🦅
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
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
1) is because for x < 0, d/dx(ln(-x)) = -1/-x = 1/x. Obviously d/dx ln(x) = 1/x for x > 0 too. Combining the two gives:
d/dx ln|x| = 1/x (for x real and non-zero). Hence integrating 1/x gives ln|x|, this is important especially if the domain of x is not restricted to the positive reals only, but also includes negative numbers
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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!