Recent content by Chenkel

That helps a lot in me understanding, thank you!

Hello everyone, I've been trying to understand the proof for the binomial theorem and have been using this inductive proof for understanding. So far the proof seems consistent everywhere it's explicit with the pattern it states, but I've started wondering if I actually fully grock it because I...

So ##\int_a^b f(u(x))u'(x)dx = \int_{u(a)}^{u(b)} f(t)dt## is still true when ##u(a) = u(b)##?

The fact that the proof you provided for the change of variables definite integration equation doesn't work with a non-injective ##u## function doesn't diminish the validity of the proof itself, right?

I agree that the left hand side of that equation doesn't equal the right side, and I think I see the general problem, if ##u=\sin \theta## and the limits of integration are ##0## to ##2\pi## then the result will be null in the change of variables formula regardless of what the original function is.

I'm not sure what you mean by somethings going wrong there. The integral ##\int_0^{2\pi} \sin \theta \ d\theta## is 0 using the fundamental theorem of calculus ##[-cos(\theta)]\Bigg|_0^{2\pi}=-\cos(2\pi) + cos(0) = 0## and the right hand side of the equation you posted is the integral from...

Interesting, thank you for the insight!

I think I probably said it right, I hope so.

Now I'm a little confused and not sure of I said that right

So perhaps the reason my proof worked is because when I let ##u=r\cos(\theta)## I kept ##\theta## in the interval ##[0, \pi]## and in that interval u is one-to-one and and invertible if I'm not mistaken.

What do we do if ##a## and ##b## are largely separated and ##u## is a periodic function that oscillates more than once in the interval ##[a, b]##? For example I chose ##u=r\cos(\theta)## and I chose ##\theta = [\pi, 0]## and I got the correct answer for the area of the circle, but if I chose...

Using what you wrote I'm able to see the how the left approach derivative of the accumulation function with the mean value theorem is equal to the function in the integrand. Thank you for the insight!

Hello everyone, I've been brushing up on some calculus and had some new questions come to mind. I notice that most proofs of the fundamental theorem of calculus (the one stating the derivative of the accumulation function of f is equal to f itself) only use a limit where the derivative is...

Hello everyone, I am curious, suppose you have a function ##f(x)=x^3## and you to find the area under the curve from 0 to x, the area would be ##\frac {x^4}{4}## but this is units of ##L^4## if x is length, but area is units of ##L^2## so what is going on here? The reason I'm curious is I...

So just to prove my formulas are consistent with the change of variables formula you proved. ##f(x) = \sqrt{r^2 - x^2}## ##u(\theta) = r\cos(\theta)## ##u'(\theta) = -r\sin(\theta)## If you expand the following equation with the previous variables it's the same equation I used in my proof...