# Differentiation/Integration Problem

• Aug 30th 2008, 10:03 PM
Mercury940
Differentiation/Integration Problem
Hello everyone,

I want to first say thank you for taking a second to look at this. My question is, on my Calc 2 homework, my professor has asked me:

given F(x) = integral of 1/t from 1 to x, x greater than or equal to 1.

First show that that derivative of F(x) (the integral being from 1 to x) = 1/x without referencing the fundamental theorm of calculus. He suggested using the mean value theorm of integration, which makes sense, but I feel like I am just rewriting the proof I looked up, which is bad because it means I don't actuallty understand what is going on. If someone can kind of explain this because my book is kind of vague about it and my professor expects us to just understand.

Second, show that the derivative of F(ax) (the integral from 1 to x) = 1/(x). I understood in Calculus 1 that this could happen since you integrate and then derive which eliminates the constant created, but I guess I can't figure how that works in the proof. Which is bad again since I don't really understand what is happening behind the scenes

Finally, show that F(ax) = F(x) - some constant C, and find the value of that constant. This one I can sort of see is a bridging of the first and second tasks but without understanding them, I can't quite grasp the entire picture. Any explanations of these without using the fundamental theorem of calculus or assuming that the anti-derivative of 1/x = ln x and the reverse, would be much appreciated.

Thank you again for your time,
Robert
• Aug 31st 2008, 12:30 AM
wingless
OK then. First we should state the MVT (mean value theorem) for integrals.

Let f be a function continous on [a,b] and differentiable on (a,b). There exists such c that satisfies,
$\displaystyle f(c) = \frac{\int_a^b f(x)~dx}{b-a}$

Now we can work on the problem.

We know that $\displaystyle F(x) = \int_1^x \frac{1}{t}~dt$. We have to find $\displaystyle \frac{d}{dx}F(x)$. Let $\displaystyle I = \frac{d}{dx}F(x)$

By the definition of derivative, $\displaystyle I = \lim_{h\to 0}\frac{\int_1^{x+h}\frac{1}{t}~dt - \int_1^{x}\frac{1}{t}~dt}{h}$

$\displaystyle I = \lim_{h\to 0}\frac{\int_x^{x+h}\frac{1}{t}~dt}{h}$

As $\displaystyle h\to h-x$,
$\displaystyle I = \lim_{h\to x}\color{red}\frac{\int_x^{h}\frac{1}{t}~dt}{h-x}$

Oh, we can use the MVT on the expression in red =)

$\displaystyle I = \lim_{h\to x} \frac{1}{c}~,~ c\in (x,h)$

As h approaches x, this limit will go to $\displaystyle \boxed {I = \frac{1}{x}}$.

Feel free to ask anything that wasn't clear to you.

Good luck on your second question (Sun)