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A problem with evaluating an integral

  1. Oct 7, 2011 #1
    [itex]\int^{\pi}_{0} f(x) dx[/itex] where ,

    f(x) = sin x if [itex]0 \leq x < \frac{\pi}{2}[/itex]
    and f(x) = cos(x) if [itex]\frac{\pi}{2} \leq x \leq \pi[/itex]

    The problem is that the version I am using of Fundamental theorem is if f is continuous on some closed interval , I wrote the integral as

    [itex]\int^{\pi / 2}_{0} f(x) dx + \int^{\pi}_{\pi /2} f(x) dx[/itex]

    but I have in the first integral f still is not continuous on [itex][0,\pi/2][/itex]

    I tried to open some references and reached another version for the theorem there f is integrable on f , and g' =f , but I couldn't do any thing

    Thanks
     
  2. jcsd
  3. Oct 7, 2011 #2

    Erland

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    Science Advisor

    That's no problem. This integral is the same as the integral of sin x over all of [0,pi/2], since the two functions are equal except in a single point. And if you change an integrable function in a single point (or finitely many points) then the function is still integrable with the same integral. This can be seen by using upper and lower sums.
     
  4. Oct 8, 2011 #3
    The problem is that the text book I am using doesn't introduce the integrals by first introducing the upper Darboux sums and the lower Darboux sum , it uses another approach which introduce the area as the limit of sums of approximating rectangles.
     
  5. Oct 8, 2011 #4
    Thanks , Now I am reading in another textbook Which deals with this point precisely .
     
  6. Oct 11, 2011 #5

    now I justified my solution as following , the first definite integral exist because f is integrable on [0,pi] (since it is piece-wise continuous ) , now since f is a function on [0,pi\2] where f= sin (x) ,0 ≤x≤ pi/2 , except at pi/2 , their definite integral is the same and then we can use the FTC , or I can use the version of the theorem which apply to any integrable function which is not necessary continuous we know that f is integrable on [0,pi/2] so we want to find a continuous function g such that g' = f on (0,pi/2) and g is continuous on [0,pi/2] so we see that this is g is sin (x) , ( I think all of this will be easy in practice ) , but what I am wandering of is why this is an exercise in my text-book , although it doesn't concentrate at this point yet .

    Thanks
     
    Last edited: Oct 11, 2011
  7. Oct 11, 2011 #6

    Erland

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    Yes, use FTC to sin x over [0,pi/2] and observe that the given function has the same integral over this interval since it differs only at one point.

    I think your textbook should have mentioned thia problem,
     
  8. Oct 11, 2011 #7

    no , it hasn't mentioned it yet , it uses a definition of integral which depends on dividing the interval into sub intervals equal in length , so I began to read a bout the theory of integral in Advanced Calculus by agnus Taylor is this good for me .
     
    Last edited: Oct 11, 2011
  9. Oct 21, 2011 #8
    So after I have read Integration it some Advanced Calculus books , I had only one question to ask , this only to confidence myself with only a part of of a proof ( you Know that the proof in Advanced books depend leaving details for you ) , it wrote that if f is integrable , then there exist partitions P' and P'' , such that given ε >0
    U(f,P') < U(f) + ε/2 and L(f,P'') > L(f) -ε/2

    Here I began to justify this to my self as following , I would only now justify the first the second one is similar , we will assume it is not true, then we have

    U(f,P') ≥ U(f) + ε/2 > U(f)

    and this contradicts that U(f) is the grates lower bound of of all Darboux sums
    But I see that this justification is not good at all because of there is P' such that U(f,P') = U(f) + ε/2 , so I began to make another one which I see is a better one assume that
    U(f,P') > U(f) + ε/4 , then this gives a contradiction , so We have

    U(f,P') ≤ U(f) + ε/4
    and since ε/4 < ε/2
    so we have U(f,P') ≤ U(f) + ε/4 < U(f) + ε/2

    Is all of What I have said is true or not



    Thanks
     
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