Hello everyone! My name is Wilson🎙️. I make videos (and animations) to help people learn math more easily 💯! I want to use my free time to make a lot of math videos to help others and to make education available to the world. Your Likes and Subscribing will help this channel be reached by more people. Please 📢 share my videos to others! Thank you! 🙏🏼
Leave me a comment as well!
The first video was published on 01/30/2020. 500 subscribers on 12/22/2020 1000 subscribers on 05/11/2021 2000 subscribers on 01/15/2022 3000 subscribers on 07/15/2022 4000 subscribers on 11/30/2022 5000 subscribers on 03/30/2023 6000 subscribers on 08/18/2023 7000 subscribers on 12/15/2023 8000 subscribers on 04/03/2023
Thank you sir. You make it kinda easier to grab the point and understand. Actually it is in my 12 class Sallybus and am trying to lesson to your video and understand. And i would be very happy if i could directly tell you my problems of Mathematics. Kindly do me a favor in this case.❤
I'm still working on knowing or rather, recognizing when you 'need' to change the order. As well, still working on when to use U sub, esp, in Polar Integrals. I guess I need to do a lot more , plenty of examples to get used to 'seeing' and 'recognizing' the problems.
I agree with you. There are certain integrands that we won't be able to integrate easily. That's when we want to change the order. With more practice, I am sure you will recognize that kind of situation and what methods to use💪
Very good video, well explained. However, you mentioned, 'I will be making another video to solve this same problem without using the formula!' but I can't find it.
If anyone is planning to use this without a square root in the function, just remember that the domain of the integrand and the answer need to be identical, in this video it is but in many cases it's not Since you introduce a square root in the final answer the domain ends up being the x values that make that sqrt argument positive or 0, which might not be the same as the domain of the integrand.
@@GlassofNumbersI see your point but the domain of 1/(2sqrt(x)) is smaller than it's antiderivative (i.e it doesn't include 0), what I'm talking about is only when the *antiderivative* ends up with a reduced domain, usually by including a square root, so I guess I should have said was to make sure the antiderivative ends up with an identical or greater domain instead of saying an identical domain. If you want a specific example try integrating 1/(4 - (x^2)) by setting x = 2sinx.
Helpful. Considering the original description of length of an arc. The right triangle... but the sqr.rt. of 1 + dy/dx doesn't make sense if you think of y on the left as the sum instead of using v as a sum of micro-hypoteneuses... so, the length of x^2 from. 0 to 1/2 cannot yet be 1... which any sum of 1 + dy/dx will be, even if it's 0. So, h= sq.rt. of x^2 + y^2... therefore, the length of x^2 from. 0 to 1/2 is closer to 1/2 + 1/16... if x=1/2 in y=x^2 then y=1/4... and 1/4^2 = 1/16.... helpful... but I still wonder how to arrive at the ln integral... someone wrote that 1/2(x^2) can be written as 1/(2x^2)... so that may be the cause.
