Another quick question, Fundamental Theorem of Calculus

[c.teixeira]

I am guessing the fundamental theorem of calculus, isn't not valid, if the integrand f depends on x. Right?

For example if he had:

$\int^{x}_{0} f(u) ( x-u) du$. If one would make F(x) = $\int^{x}_{0} g(u) du$, with g(u) = f(u) ( x-u). Then F`(x) = g(x) = f(x) (x-x) = 0. But this is not correct as you know. He have to get the x out of the integral.

So, concluding, The First Fudamental Theorem of Calculus, is valid only if the integrand f , doens't not depend on x. Correct?

 

[Millennial]

No. If x is a constant, it is treated as one in the integral. There is nothing special about x, it's just that it is the upper limit of integration. That does not change how the integration is carried out. Namely, your integral converts to
$$\int_{0}^{x}xf(u)-uf(u)\,du=x\int_{0}^{x}f(u)du-\int_{0}^{x}uf(u)du$$
to which the FFTC is applicable.

 

[c.teixeira]

No. If x is a constant, it is treated as one in the integral. There is nothing special about x, it's just that it is the upper limit of integration. That does not change how the integration is carried out. Namely, your integral converts to
$$\int_{0}^{x}xf(u)-uf(u)\,du=x\int_{0}^{x}f(u)du-\int_{0}^{x}uf(u)du$$
to which the FFTC is applicable.

First of all thank you! But I didn't quite understood your explanation.

Yes, I understand that that is the correct way to differentiate the expression. In fact that was exactly why I got this doubt to begin with.


When I saw:
$\int^{x}_{0}f(u)(x-u)$ = $$\int_{0}^{x}xf(u)-uf(u)\,du=x\int_{0}^{x}f(u)du-\int_{0}^{x}uf(u)du$$

I tought:

They "expanded" f(u) (x-u) to f(u)x - f(u)u, so that they could differentiate using the FFTC and the rules for the differentiation of the product of two fuctions namely l(x)=x and F(x) = $\int^{x}_{0}f(u)du$.

That is why, I got the idea that one couln't use the FFTC to differentiate direcltly a function F(x) = $\int^{x}_{0}g(u)du$., with in which g(u) depends on x, altought it is perceived as a constant during integration.

To summarize my idea, I guess the following is the best way:

Let g(u) = f(u)(x-u).

and F(x) = $\int^{x}_{0}g(u)du$.

Then throught direct apllication of the FFTC we would get F`(x) = g(x)=f(x)(x-x) = 0.,

Which is obvsiouly different from $$\int_{0}^{x}xf(u)-uf(u)\,du=x\int_{0}^{x}f(u)du-\int_{0}^{x}uf(u)du$$= $\int^{x}_{0}f(u) du$ + xf(x) - xf(x).

What brings me to the same question:

Is or isn't the FFTC applicable when the integrand f in F(x) = $\int^{x}_{0}f(u)du$, depends on the constant x.

If is applicable to such case, why is it wrong in my example to assume F`(x) = g(x) = f(x) (x-x) = 0.

Regards,

c.teixeira

 

[Millennial]

"Then throught direct apllication of the FFTC we would get F`(x) = g(x)=f(x)(x-x) = 0.,"

This is false. How do you know g(0)=0?

 

[c.teixeira]

"Then throught direct apllication of the FFTC we would get F`(x) = g(x)=f(x)(x-x) = 0.,"

This is false. How do you know g(0)=0?

I don't know that, but (x-x) would be zero, making g(0) = 0. right?

 

[qbert]

you can't *directly* use the FTC in the case you mention.

the idea your skirting is called leibniz rule
http://en.wikipedia.org/wiki/Leibniz_integral_rule

it is just an application of the FTC and the chain rule.
$$\frac{d}{dx} \int_0^x f(u,x) du = f(x,x) + \int_0^x \frac{\partial f}{\partial x}(u,x) du$$

 

[c.teixeira]

you can't *directly* use the FTC in the case you mention.

the idea your skirting is called leibniz rule
http://en.wikipedia.org/wiki/Leibniz_integral_rule

it is just an application of the FTC and the chain rule.
$$\frac{d}{dx} \int_0^x f(u,x) du = f(x,x) + \int_0^x \frac{\partial f}{\partial x}(u,x) du$$

Thanks!