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Solving for a derivative of an integral

  1. Aug 7, 2006 #1

    dnt

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    the question is find a function, f, and a value, a, such that:

    2 times the integral from a to x of f(t)dt = 2sinx - 1

    (sorry i dont know how to make it look nice)

    well the first thing i did was divide by 2, then take the derivative of each side:

    f(x) - f(a) = cos x

    and now im stuck. am i correct so far? where do i go from here? thanks.
     
  2. jcsd
  3. Aug 7, 2006 #2
    You made a mistake.
    If you took the derivative of both sides dx, why would F(a) be differentiated to f(a)? Wouldn't it be 0?
     
  4. Aug 7, 2006 #3
    Why did you divide by 2, and what happened to that factor of 2 when you took the derivative? You also might want to recheck your work taking the derivative, do remember what the second fundamental theorem of calculus says?
     
  5. Aug 7, 2006 #4

    dnt

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    i thought it would be the derivative of f at a (whatever value that may be...i dont know because i dont know what a is)
     
  6. Aug 7, 2006 #5

    dnt

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    i thought i should divide by 2 just to get rid of it. on the right side it becomes sin x - 1/2. after i take the derivative of that i get cos x.
     
  7. Aug 7, 2006 #6
    Yes, but what you do to one side of the equation you must do to the other side as well and so I asked what happened to the factor of 1/2 that would have been multiplying the integral?

    [tex]\frac{d}{dx}\int_{a}^{x} f(x) dx\ = \ f(x) [/tex]

    That is very basically what the second fundamental theorem of calculus says and is what you need to use in this problem.
     
  8. Aug 7, 2006 #7
    Nope.
    Think about it this way...
    A definite integral yields a numerical answer... UNLESS one of the limits of integration is a variable that you're dealing with-- could be x OR a.
    In this case, you have taken the derivative of both sides with respect to x-- thus treating x as the variable, not a...
    In any case, F(a) does not have anything to do with x... differentiating it just kills it; thus, it disppears...
     
  9. Aug 7, 2006 #8

    dnt

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    i guess i dont get how you take the derivative of an integral. the integral from say, a to x, is the area under the curve, right? what does the derivative of an area mean?
     
  10. Aug 7, 2006 #9
    Well, according to the fundamental theorem of calculus, derivative and integrals are opposites..
    Say you have a function f..
    Take the integral, you have a function F.
    Differentiate F, you get f again!
    That's it, really...
     
  11. Aug 7, 2006 #10

    dnt

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    right, so then when i take the derivative of the integral from a to x of f(t) dt, shouldnt it just be f(x) - f(a)?
     
  12. Aug 7, 2006 #11

    dnt

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    and i still dont get how im suppose to take the derivative of a function when i dont even know what it is.

    FTC has always been a very tough topic for me to comprehend. still trying to understand all of this.
     
  13. Aug 7, 2006 #12
    If we define the integral of f(x) to be F(x) then if we take the integral from a to x we get F(x) - F(a), but when we differentiate with respect to x we treat a as a constant so the derivative of that integral will be F'(x) which is equal to f(x) because of how we defined F(x), F(a) will just be some consatnt that when we differentiate just disappears since it will be zero. Take a look at my previous post for a rather informal statement of the second fundamental theorem of calculus.
     
  14. Aug 10, 2006 #13
    A worked solution

    The attached document has a worked solution to your problem. The fundamental theorem of calculus makes more sense if you think of integrals as anti-derivatives, as they were first taught to me.
     

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