The fundamental theorem of calculus(I think;) )

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SUMMARY

The discussion centers on applying the Fundamental Theorem of Calculus to differentiate the integral function ∫1x2 ln(t2) dt. Participants clarify that to differentiate this integral with respect to x, one must use the chain rule due to the upper limit being x2 instead of x. The correct approach involves substituting u = x2 and applying the chain rule, leading to the derivative expression df/dx = (df/du)(du/dx). The discussion emphasizes that the differentiation must align with the variable present in the upper limit of integration.

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beaf123
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Homework Statement



Been doing some old exams lately and found out that something I have problems with is questions of the type ( example):
Differente the function:

∫ (x^2 ),(1), ln(t^2) dt

Sorry for the bad writing.
(x^2 ),(1), is the intgral from 1 to X^2

It should be fairly simple, but my brain somehow clicks when I see this question. Maybe because its integration AND derrivation. Could someone explain the steps they use to solve this exircise.

And would I get the same answer if I integrated the function ( maybe not the best to do in this case ), solved for the integral and then differented the integral?

Thank you!:)
 
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So your problem is "Find the derivative of \int_1^{x^2} ln(t^2)dt".

You mention the "Fundamental Theorem" and I assume you mean the "Fundamental Theorem of Calculus"! One part of that says that d/dx \int_a^x f(t)dt= f(x)
Here, the only problem is that the upper limit is x^2 rather than x.

So let u= x^2. The integral becomes \int_1^u ln(t^2)dt. What is the derivative of that with respect to u? And to find the derivative with respect to x, use the chain rule: df/dx= (df/du)(du/dx).
 
Im confused. Why does I have to use the chain rule? Cant I just stick X^2 in there and get: ln(X^2)^2 = ln (x)^4 = 4lnx?

;-)
 
Another way to think about it is that after doing the integration, you will have some expression like F(x^2)-F(1). Now, when you take the derivative, you will have to use the chain rule (and remember F(1) is a constant) so that
\frac{d}{dx}(F(x)-F(1)) = F'(x^2)\frac{d(x^2)}{dx}. Now, use the fact that F'=f (which is the FTC you reference).
 
beaf123 said:
Im confused.
Yes, very much.
beaf123 said:
Why does I have to use the chain rule? Cant I just stick X^2 in there and get: ln(X^2)^2 = ln (x)^4 = 4lnx?
It's hard to know where to start.

Are you thinking that
$$ \int ln(t^2) dt = (ln(t^2))^2?$$

To use the Fundamental Theorem of Calculus, all of the important pieces have to be exactly as in the theorem -- differentiation has to be with respect to the same variable as is present in the upper limit of integration.

For your problem, it's differentiation with respect to x, but it's x2 as a limit. Reread what HallsOfIvy said you need to do.
 
beaf123 said:

Homework Statement



Been doing some old exams lately and found out that something I have problems with is questions of the type ( example):
Differente the function:

∫ (x^2 ),(1), ln(t^2) dt

Sorry for the bad writing.
(x^2 ),(1), is the intgral from 1 to X^2

It should be fairly simple, but my brain somehow clicks when I see this question. Maybe because its integration AND derrivation. Could someone explain the steps they use to solve this exircise.

And would I get the same answer if I integrated the function ( maybe not the best to do in this case ), solved for the integral and then differented the integral?

Thank you!:)

You can write this as ∫_{1..x^2} ln(t^2) dt, or as ∫{ln(t^2): t=1..x^2}. Anyway, do you know the general formula for
\frac{d}{dx} \int_1^{f(x)} F(t) \, dt? If you do, just use it. Note that you can find the derivative of the integral, even if you cannot do the integral itself.

RGV
 
Im going to spend some time working with this, but
if we had the same exircise but with x instead of X^2, would the answer be ln(x^2)?
 
beaf123 said:
Im going to spend some time working with this, but
if we had the same exircise but with x instead of X^2, would the answer be ln(x^2)?
Yes.
 

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