Examples of disc method for the volume of the solid of revolution: kzread.info/dash/bejne/p515p6exZ9C2eMo.htmlsi=Znn3zTxHzseDBjQP how to use these formulas: kzread.info/dash/bejne/moFqvNNtlpmeqbg.html

This is so brilliant. So many students could learn from this, instead of just blindly applying a formula, instead understand the how's and why's and derive the formulas themselves when needed.

Its phenomenal how good you explain this formulars.

@bprpcalculusbasics

2 ай бұрын

Thank you!

Love how you use integrals to calculate volume and area. In my secondary school years, it was taught that integrals were jnvented for this case, and the teacher used all this in several sessions to explain the why and what of integrals.

@thexoxob9448

15 күн бұрын

Integrals mean the sum as dx gets really small

I love how dedicated you are in making educational content that makes it accessible to the general public. Long may it lasts ⭐⭐⭐⭐⭐

Hey, this is really great! I saw that the ordinary math basics channel is really quite basic at times, not dismissing even basic operations like addition or multiplication as super obvious. this calculus basics channel doesn't seem to have as much of the same true basics, on integrals and derivatives and such. Hopefully there will be more of the "you did a few years of calculus in high school but forgot all about it" types of primer videos over time!

Thank you so much man. I have consumed hours of your content, and it has taught me so much math, and inspired me to get even better at math and learn advanced concepts; which has been made so much easier through your content. Thank you so much for preparing me for the difficult classes in high school, even though I am in the seventh grade, and I wish you the best and only the best.

@bprpcalculusbasics

2 ай бұрын

Happy to help! Thank you!

Proof of formula at 17:44 We know area of sector is ½r²θ, and circular arc length is rθ, where r is radius and θ is angle of sector. Suppose we let r be the smaller radius, and R be the larger radius Then, l = middle radius × θ = ½(R + r)θ area of the "strip" = ½R²θ - ½r²θ = ½(R² - r²)θ = [½(R + r)θ] × (R - r) = lw

@salmonsushi47

17 күн бұрын

not all the cut out strips for different functions would be of that shape right? so isnt it better to go with a general limit intuition that as we reduce the width of strip cut outs approximate rectangle more and more?

Worlds best prof.

One of the application of integration in calculus is to paint an area of an object regardless of the shape. If you try to do this in code, you can try iterate over a set of parameter to a function, and draw a single line every time. This is what i tried to do in my app.later i will try to fix it by using unit testing.

Outstanding!

Simply beautiful,Sir...Thank You very much...

Excellent

the last shape ,which you called a part of a cone ,in india we studied that as frustum of a cone with c.s.a pi.l(r1 + r2),hey just telling

@ *bprp calculus basics* -- It would be good for the audience for you to demonstrate the same example across each of area, volume, arc length, and surface area to make it more concrete.

@bprpcalculusbasics

2 ай бұрын

Yes. I have the example videos in the pinned comment.

Hi, thanks for the video! how can we formally proof these formulas? Because this is only a geometric rappresentation of the situation

@kristopherwilson506

2 ай бұрын

A real analysis class :) formally showing these can be complicated. Since integration is a limiting process, we know that the distance between what we want-in this case, the Riemann sum and the value of the integral-needs to be less than some arbitrarily small value epsilon.

About the S.A. part don't you jave to add the areas of the circles?

I think it is good to publish a book that contains all the formulas

Lovely

Saw you hesitate a bit as to whether to spell the word dis(c/k) with a c or a k. Strong disagreement between the manufacturers of floppies, the manufacturers of CDs, and spinal surgeons.

@bprpcalculusbasics

2 ай бұрын

Good catch! 😆

@lostwizard

2 ай бұрын

General rule of thumb: use "disc" unless it is referring to a rotational magnetic storage medium or something using the same form factor. (Floppy disk, hard disk (even solid state), but compact disc, spinning disc, the disc of the sun, etc.)

Can you please do one for calc 3 integrals? The double integral, line integral & surface integral

@bprpcalculusbasics

2 ай бұрын

Unfortunately I am not too familiar with those topics since I haven’t taught it, but hopefully one day!

You can think that "((dy)/(dx))^2" is the same thing as "(f'(x))^2". Great video though!

@thewok3576

2 ай бұрын

You can also integrate the circumference of a disc in order to get the surface area of any volumetric object.

@thewok3576

2 ай бұрын

I don't think dL is necessary, dx works as well (if I'm not mistaken).

Integral of ((1-x^7)^(1/4) - (1-x^4)^(1/7)) can you please solve this integral BPRP?

@cyrusyeung8096

2 ай бұрын

Wolfram Alpha says you would need the hypergeometric function, so it is non-elementary.

@bprpcalculusbasics

2 ай бұрын

I think you wanted it from 0 to 1?

@sinekavi

2 ай бұрын

@@bprpcalculusbasics Yes from 0 to 1

@josip.harasic

2 ай бұрын

=0

Coooool

Hello

@bprpcalculusbasics

2 ай бұрын

Hi!

ds for arclength, dS for surface area, so you don't mix them up. (I've also seen dσ for surface area so that dS can be used for the vector version.) But really, ds for arclength is a very strange letter to use!

@carultch

2 ай бұрын

Also, draw a cursive s to tell it apart from a 5, and to tell lowercase s apart from capital S.

How to proof that the definite integral is F(b) - F(a) ?

@naitikshukla2020

28 күн бұрын

Same doubt

There's a lot of abuse of notation But i still like it

@bprpcalculusbasics

2 ай бұрын

That’s why I didn’t say I was proving these formula in the video haha.

Also, these all come from double/triple integrals: A = ∬dA = ∬dydx= ∬rdrdθ = ∫ydx = 0.5∫r^2dθ V = ∭dV = ∭dzdydx = ∭rdzdrdθ = ∬zdA = ∬zdydx= ∬zrdrdθ = ∫Adx = ∭rdrdθdx = 0.5∬r^2dθdx = π∫r^2dx L = ∮u ∙ dr = ∫ds = ∫√(1+y'^2)dx = ∫√(x'^2+y'^2)dt = ∫√(r^2+r'^2)dθ SA = ∯u ∙ dS = ∬dS = ∬√(1+z_x^2+z_y^2)dydx = = ∫2πρds

Hi, my math teacher could not solve this, if someone could help me please xd 5^x+3^x=7

@sinekavi

2 ай бұрын

Take natural log on both sides...... ln5^x + ln3^x=ln7 xln5 + xln3=ln7 x(ln5+ln3)=ln7 x=ln7/(ln5+ln3) I am not sure whether I am correct...... So please check with your teacher again whether my answer is correct

@blansimon7227

2 ай бұрын

​@@sinekaviWhen we take natural logarithm on both sides the whole expression is caged, so in the first step should be: Ln(5^x+3^x)=Ln7 So it's not correct, but thanks for trying anyway!