Homework question involving 2nd FTC, derivative of an integral

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Homework Help Overview

The discussion revolves around the application of the second Fundamental Theorem of Calculus (FTC) to find the second derivative of a function defined by an integral involving a differentiable function f. The integral is taken from -x to x of the sum of f(t) and f(-t).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss breaking down the integral into two parts and applying the second FTC to find H'(x). There are questions about the implications of differentiating f(-t) and whether the chain rule should be applied. Some participants express uncertainty about the effects of the negative sign in the limits of integration.

Discussion Status

Several participants have provided their attempts at deriving H'(x) and are now considering how to proceed with finding H''(x). There is a recognition of the need to apply the chain rule when differentiating terms involving f(-t). The conversation reflects a collaborative exploration of the problem without reaching a definitive conclusion.

Contextual Notes

Participants are navigating the complexities of differentiating integrals with variable limits and the implications of the function's symmetry. There is an acknowledgment of the need for clarity on the application of the chain rule in this context.

mrchris
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Homework Statement


Suppose that the function f : R → R is differentiable. Define the function H: R → R by
H(x) = ∫−x to x of [ f (t) + f (−t)]dt for all x in R.
Find H'' (x).

Homework Equations





The Attempt at a Solution


i divided it up into ∫ 0 to x of [ f (t) + f (−t)]dt and -∫ from 0 to -x of [ f (t) + f (−t)]dt. then by the 2nd FTC, H '(x)=
2[ f (x) + f (−x)], so H ''(x) should then be of 2[ f '(x) - f '(−x)]. I'm just not sure if the f(-t) changes anything besides the fact that when I differentiate f(-t) I use the chain rule and multiply by -1.
 
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mrchris said:

Homework Statement


Suppose that the function f : R → R is differentiable. Define the function H: R → R by
H(x) = ∫−x to x of [ f (t) + f (−t)]dt for all x in R.
Find H'' (x).

Homework Equations





The Attempt at a Solution


i divided it up into ∫ 0 to x of [ f (t) + f (−t)]dt and -∫ from 0 to -x of [ f (t) + f (−t)]dt.
Looks OK up to here.

So $$ H(x) = -\int_0^{-x}(f(t) + f(-t))dt + \int_0^{x}(f(t) + f(-t))dt $$

Show me how you got H'(x).
mrchris said:
then by the 2nd FTC, H '(x)=
2[ f (x) + f (−x)], so H ''(x) should then be of 2[ f '(x) - f '(−x)]. I'm just not sure if the f(-t) changes anything besides the fact that when I differentiate f(-t) I use the chain rule and multiply by -1.
 
so i am under the impression that d/dx of ∫f(t) dt = f(x), so taking each term, H'(x)= dH/dx of
∫0 to x [f(t)]= f(x)
∫0 to x [f(-t)]= f(-x)
∫0 to -x [f(t)]= f(-x)*-1
∫0 to -x [f(-t)]= f(x)*-1

so H'(x)= [f(x)+f(-x)]-(-1)[f(-x)+f(x)]=[f(x)+f(-x)]+[f(-x)+f(x)]=2[f(x)+f(-x)]

I am not sure if I am also supposed to be applying the chain rule at this step with the f(-t) derivative
 
This looks fine. Now take the derivative again, keeping in mind that d/dx(f(-x)) = -f'(-x).
 
so H ''(x) should then be of 2[ f '(x) - f '(−x)].
 
mrchris said:
so i am under the impression that d/dx of ∫f(t) dt = f(x), so
You mean, I hope, that [itex]d/dx \int_a^x f(t)dt[/itex] so that [itex]d/dx \int_x^a f(t)dt= d/dx\left(-\int_a^x f(t)dt\right)= -f(x)[/itex]

taking each term, H'(x)= dH/dx of
∫0 to x [f(t)]= f(x)
∫0 to x [f(-t)]= f(-x)
∫0 to -x [f(t)]= f(-x)*-1
∫0 to -x [f(-t)]= f(x)*-1

so H'(x)= [f(x)+f(-x)]-(-1)[f(-x)+f(x)]=[f(x)+f(-x)]+[f(-x)+f(x)]=2[f(x)+f(-x)]

I am not sure if I am also supposed to be applying the chain rule at this step with the f(-t) derivative
 

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