What step have you taken to move the material to Africa, because the youtube thing and web apps are Luxurious in some way here. Am from Pretoria, South Africa and really gained a lot from Khan's presentations. Thanks.
I have a question: When you're taking cross sections of a solid, does the height have to be proportional to the base or can the height remain constant? For example, if the cross sections were squares could the volume just be the integral of 2s (if 2 is the height) instead of s^2?
Sorry Aretwo, that would not give the correct solution. f(x)^2 - g(x)^2 does not equal (f(x) - g(x))^2, and neither of those are equal to (1/4)*(f(x) - g(x))^2 [area of a single triangular cross-section]
These videos save my life every single day.
Your drawings are a world of help! Up until this point I was clueless as to how to visualize these shapes. Many thanks!
Seriously the best channel for calculus tutorials. Your explanations earned me an A in Calc 1, and I'm hoping they do the same for 2 and 3. Thank you!
Wow thank you my friend I didn't really understand this in class but now it makes sense
Nice refresher of Calculus II basics. Great Video and commentary
What step have you taken to move the material to Africa, because the youtube thing and web apps are Luxurious in some way here. Am from Pretoria, South Africa and really gained a lot from Khan's presentations. Thanks.
thank u mr.khan u are the best
Thanks king
my eyes have been opened
In Sweden we call this the "slice-method" :)
massivejester That sounds so much more slick
same in india as well
Im still not sure if i got all that
I dont even understand shit!! I just watch it because it looks interesting ... So weird
+EvryShotKillz29 what grade are you in rn?
nice
why is it a right triangle
Looks like this thread is averaging about 3 comments per year.
I have a question: When you're taking cross sections of a solid, does the height have to be proportional to the base or can the height remain constant? For example, if the cross sections were squares could the volume just be the integral of 2s (if 2 is the height) instead of s^2?
Why would you call the base of the isosceles triangle "h". So confusing and hard to follow.
I don't understand the point of doing that when you can simply find the integral of f(x)^2-g(x)^2.
help picture what you're doing.
That would only give you the area. Looking for volume here
Sorry Aretwo, that would not give the correct solution.
f(x)^2 - g(x)^2 does not equal
(f(x) - g(x))^2, and neither of those are equal to
(1/4)*(f(x) - g(x))^2 [area of a single triangular cross-section]
Only 18 comments lol
Only 12 subscribers lol !