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Fundamental Theorem

  • Thread starter irok
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  • #1
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Homework Statement
Question One:
Find a continuous function f and a number a such that

[tex]2 + \int_{a}^{x} \frac {f(t)} {t^{6}} \,dt = 6 x^{-1}[/tex]

Question Two:
At what value of x does the local max of f(x) occur?
[tex]f(x) = \int_0^x \frac{ t^2 - 25 }{ 1+\cos^2(t)} dt[/tex]

The attempt at a solution
I just need some pointers of where to get started.
Question One:

So I used FTC1 on both sides,

[tex]2 + f(x) / x^{6} = 6x^{-1}[/tex]

[tex]f(x)= 6x^{5} - 2[/tex]

I'm not sure how to find a, evaluation theorem?

Question Two:
 
Last edited:

Answers and Replies

  • #2
I don't know about an analytic solution, but the second part of the problem is very feasible numerically. You can solve in Mathematica in only a few lines by turning it into a minimization problem.
 
  • #3
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Well, for Question one:

Can anyone confirm that [tex]f(x) = 6x^{5}[/tex] and a = 2.

I'm pretty sure that a = 2 since,

F(x) - F(a) = [ [tex]6x^{5} / x^{6}[/tex] ] - [ 2 ] = [tex]6x^{-1} - 2[/tex]
 
  • #4
matt grime
Science Advisor
Homework Helper
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How does the FTC just let you drop an integral sign out like that? (In 1.)
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement
Question One:
Find a continuous function f and a number a such that

[tex]2 + \int_{a}^{x} \frac {f(t)} {t^{6}} \,dt = 6 x^{-1}[/tex]

Question Two:
At what value of x does the local max of f(x) occur?
[tex]f(x) = \int_0^x \frac{ t^2 - 25 }{ 1+\cos^2(t)} dt[/tex]

The attempt at a solution
I just need some pointers of where to get started.
Question One:

So I used FTC1 on both sides,

[tex]2 + f(x) / x^{6} = 6x^{-1}[/tex]
No, that is not correct. You have differentiated the left side of the equation but not the right.

[tex]f(x)= 6x^{5} - 2[/tex]

I'm not sure how to find a, evaluation theorem?

Question Two:
Once you have found the function, put it into the integeral.
 

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