Derivative of Integral with Respect to Variable of Integration

[Mothrog]

I have a function of the form

\int^{r}_{0} tf(t)dt

I'm supposed to take the derivative with respect to r of this integral. By the fundamental theorem of calculus, is the derivative not

rf(r)

The problem being I need, for the problem to work out correctly, to have a df/dr term. So, am I mistaken in my solution?

 

[sutupidmath]

well, it depends on what properties does the function under the integral sign has, tf(t).
THere are some conditions which this function has to fulfill in order for the answer to be what u said. THe first one is that f(t)>0, which implies directly that \int^{r}_{0} f(t)dt is monotono increasing.

 

[HallsofIvy]

No, there is no need that f(t)> 0. By the fundamental theorem of calculus, as long as g(x) is continuous if G(x)= \int_a^x g(t)dt, then \frac{dG}{dx}= g(x). In this case, g(x)= xf(x) so, as long as f is continuous, if F(r)= \int_0^r tf(t)dt, then \frac{dF}{dr}= rf(r). I do not understand why you would "need to have a df/dr term. Perhaps if you were to post the actual problem, it would make more sense.

 

[Mothrog]

It is desired to show that if, for constant k,

\frac{d}{dr} g(r) = \frac{d}{dr}k\frac{r^2}{\int^{r}_{0} tf(t)dt} < 0​

Then

\frac{d}{dr}f(r) > 0​

When I take the derivative of g, I get

\frac{dg}{dr} = k\frac{2r\int^{r}_{0}tf(t)dt - r^2(rf(r))}{(\int^{r}_{0} tf(t)dt)^2}​

If that is correct, I don't see how it is possible to show the desired relation. The definition of g given is the only information known about g. Am I missing something?

 

[Mothrog]

Bump.

 

[Dick]

Well, here's something. Let's throw k away. If it's positive it doesn't add anything to the problem and if it can be either sign, then it just plain isn't true. Now let

F(r)=\int^{r}_{0} tf(t)dt and G(r)=\frac{r^2}{2} f(r).

Now look at the numerator of dg/dr. If it's negative then G(r)>F(r). But G(0)=F(0)=0. So if G'(r)>F'(r) then I could conclude G(r)>F(r). G'(r)=rf(r)+(r^2/2)*f'(r). F'(r)=rf(r). So if f'(r)>0 then dg/dr<0 and vice versa. Trouble is, I can't see how the converse would hold, which seems to be what you are actually trying to prove.

 

[Mothrog]

Well, here's something. Let's throw k away. If it's positive it doesn't add anything to the problem and if it can be either sign, then it just plain isn't true. Now let

F(r)=\int^{r}_{0} tf(t)dt and G(r)=\frac{r^2}{2} f(r).

Now look at the numerator of dg/dr. If it's negative then G(r)>F(r). But G(0)=F(0)=0. So if G'(r)>F'(r) then I could conclude G(r)>F(r). G'(r)=rf(r)+(r^2/2)*f'(r). F'(r)=rf(r). So if f'(r)>0 then dg/dr<0 and vice versa. Trouble is, I can't see how the converse would hold, which seems to be what you are actually trying to prove.

I'm not sure you can assume g(0) = F(0) = 0. Can you walk me through your logic on that step?

 

[Dick]

I'm not sure you can assume g(0) = f(0) = 0. Can you walk me through your logic on that step?

Not the little f and g. The big F and G. It's not an assumption, look at their definitions.

 

[Mothrog]

Not the little f and g. The big F and G. It's not an assumption, look at their definitions.

But G(r) depends only on little f(r), so how can you conclude that if F(0) = 0, that G(r) = 0?

 

[Dick]

G(0)=f(0)*0^2/2=0. F(0) is the integral from 0 to 0 of t*f(t) which=0.

 

[Mothrog]

G(0)=f(0)*0^2/2=0. F(0) is the integral from 0 to 0 of t*f(t) which=0.

Ah, OK. That makes sense.

 

[Mothrog]

Well, here's something. Let's throw k away. If it's positive it doesn't add anything to the problem and if it can be either sign, then it just plain isn't true. Now let

F(r)=\int^{r}_{0} tf(t)dt and G(r)=\frac{r^2}{2} f(r).

Now look at the numerator of dg/dr. If it's negative then G(r)>F(r). But G(0)=F(0)=0. So if G'(r)>F'(r) then I could conclude G(r)>F(r). G'(r)=rf(r)+(r^2/2)*f'(r). F'(r)=rf(r). So if f'(r)>0 then dg/dr<0 and vice versa. Trouble is, I can't see how the converse would hold, which seems to be what you are actually trying to prove.

Actually, looking back at the problem again, I needed to show that f'(r) > 0 implies g'(r) < 0, so I don't need to prove the converse. So, problem solved. Thanks for your help. I was tearing my hair out over that one.

 

[Dick]

No problem. Good thing you don't have to prove the converse, It's not true.