1. ## Integrals

Suppose f(x) is differentiable on (−+),
F(x)=xf(t)dt

With the boundaries being from 1/x to 0.

for x/=0. then, what is F(x)?

Sorry for the sloppy notation, I'm not sure the math code for integrals and such. I'm not sure how to go about solving this question. I need to ultimately find the derivative, but first I believe I need to integrate. However, I do not know f(t), so how can I integrate it? As well, there are two different variables... So confused. Any help would be greatly appreciated...

2. Originally Posted by Hellreaver
Suppose f(x) is differentiable on (−+),
F(x)=xf(t)dt

With the boundaries being from 1/x to 0.

for x/=0. then, what is F(x)?

Sorry for the sloppy notation, I'm not sure the math code for integrals and such. I'm not sure how to go about solving this question. I need to ultimately find the derivative, but first I believe I need to integrate. However, I do not know f(t), so how can I integrate it? As well, there are two different variables... So confused. Any help would be greatly appreciated...
First off, is this the integral

$F(x) = \int_{\frac{1}{x}}^0 x f(t)\, dt$

3. That is it, except that the upper and lower bounds are switched, so 1/x is the upper bound and 0 is the lower. Thanks, I have no idea how to write the questions out like that.

4. Originally Posted by Hellreaver
Suppose f(x) is differentiable on (−+),
F(x)=xf(t)dt

With the boundaries being from 1/x to 0.

for x/=0. then, what is F(x)?

Sorry for the sloppy notation, I'm not sure the math code for integrals and such. I'm not sure how to go about solving this question. I need to ultimately find the derivative, but first I believe I need to integrate. However, I do not know f(t), so how can I integrate it? As well, there are two different variables... So confused. Any help would be greatly appreciated...
$F(x) = \int^{\frac{1}{x}}_0 x f(t)\, dt = x \int^{\frac{1}{x}}_0 f(t)\, dt = x \cdot G(x)$.

Let $w = \frac{1}{x}$: $~ G(w) = \int^{w}_0 f(t)\, dt$.

To differentiate $G(x)$ apply the chain rule and Fundamental Theorem of Calculus:

$\frac{dG}{dx} = \frac{dG}{dw} \cdot \frac{dw}{dx} = f(w) \cdot \frac{dw}{dx}$.

It is left for you to substitute back $w = \frac{1}{x}$ to get the final expression for $\frac{dG}{dx}$.

Knowing $\frac{dG}{dx}$ you can now easily use the product rule to differentiate $F(x)$.

Then differentiate the result again.

5. Originally Posted by mr fantastic
$F(x) = \int^{\frac{1}{x}}_0 x f(t)\, dt = x \int^{\frac{1}{x}}_0 f(t)\, dt = x \cdot G(x)$.

Let $w = \frac{1}{x}$: $~ G(w) = \int^{w}_0 f(t)\, dt$.

To differentiate $G(x)$ apply the chain rule and Fundamental Theorem of Calculus:

$\frac{dG}{dx} = \frac{dG}{dw} \cdot \frac{dw}{dx} = f(w) \cdot \frac{dw}{dx}$.

It is left for you to substitute back $w = \frac{1}{x}$ to get the final expression for $\frac{dG}{dx}$.

Knowing $\frac{dG}{dx}$ you can now easily use the product rule to differentiate $F(x)$.

Then differentiate the result again.
I am lost at the chain rule and Fundamental Theorem part of this. Can you elaborate on what you did here?

6. Originally Posted by Hellreaver
I am lost at the chain rule and Fundamental Theorem part of this. Can you elaborate on what you did here?
Read this: The Second Fundamental Theorem of Calculus

7. Here is a little tid bit to remember.

$\frac{d}{dx}\int_{a}^{g(x)}f(t)dt=f(g(x))g'(x)$

And:

$\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$

Now, prove each of these.

The former is what Mr F is getting at.

Hopefully, you know what the chain rule is by now?. That is one of the first things to learn in calc. It is invaluable.

8. Sorry if I'm being slow, but what is the f(w) value supposed to be? I know that w=1/x, but how do I derive the function? Is it just f'(1/x)?

9. Originally Posted by Hellreaver
Sorry if I'm being slow, but what is the f(w) value supposed to be? I know that w=1/x, but how do I derive the function? Is it just f'(1/x)?
All you do is substitute $w = \frac{1}{x}$ into the result I gave you: $\frac{dG}{dx} = f\left( \frac{1}{x}\right) \cdot \left(- \frac{1}{x^2}\right)$.

This gives you the derivative that you need when applying the product rule to find $\frac{dF}{dx}$.