Fundamental Theorem of Calculus problem

Pacopag
Messages
193
Reaction score
4

Homework Statement


The FTC states that
{d\over{dx}}\int_a^x f(t)dt = f(x)
Now, how do I do something like
{d\over{dx}}\int_a^{g(x)} f(t)dt = ?

Homework Equations


The Attempt at a Solution


I know that it has to do with the chain rule, but I forgot my textbook at school and I can't seem to find it online (e.g. Wikipedia).
 
Physics news on Phys.org
Pacopag said:

Homework Statement


The FTC states that
{d\over{dx}}\int_a^x f(t)dt = f(x)
Now, how do I do something like
{d\over{dx}}\int_a^{g(x)} f(t)dt = ?


Homework Equations





The Attempt at a Solution


I know that it has to do with the chain rule, but I forgot my textbook at school and I can't seem to find it online (e.g. Wikipedia).

Changing the variable to u, where u = g(x), the integral looks like this:
{d\over{du}}\int_a^u f(t)dt = f(u)

The trouble is, you want {d\over{dx}}\int_a^u f(t)dt = f(u)

So if you want d/dx(H(u)), that's the same as d/du(H(u))*du/dx, isn't it? (Here, H(u) represents the value of the definite integral."
 
Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top