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Math Help - Please Help with an integration Q

  1. #1
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    Please Help with an integration Q

    Hey I'm stuck on this Q and I would be grateful for any replies.

    I have tried twice to write the problem in latex but unfortunately I am not proficient enough in it.

    I have attached the Q NB it is Q2 on the attached photo
    Also I have attached the graph that I think is right for the first part
    but my real Q is for part 1 how do you show that that integral is equal to that sum? I can see it clearly from
    the graph but I don't know how to show it?
    I'll work on part two later.

    I would be really grateful for any replies.

    Thanks from James
    Attached Thumbnails Attached Thumbnails Please Help with an integration Q-058.jpg   Please Help with an integration Q-061.jpg  
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  2. #2
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    Re: Please Help with an integration Q

    First,
    \int_{0}^a\sqrt{[x]}dx=   \int_{0}^1\sqrt{[x]}dx+\int_{1}^2\sqrt{[x]}dx+\cdots +\int_{a-1}^a\sqrt{[x]}dx

    second, note that in the first interval (integral)  [x]=\sqrt{0}
    in the second interval (integral) [x]=\sqrt{1}
    and so on, unitil
    the ath integral  [x]=\sqrt{a-1}, so
    \int_{0}^a\sqrt{[x]}dx=   \int_{0}^10dx+\int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx
    =   \int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx     =1+2+\cdots+\sqrt{a-1} as desired.
    Last edited by Kmath; February 20th 2013 at 07:52 AM.
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  3. #3
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    Re: Please Help with an integration Q

    Quote Originally Posted by Kmath View Post
    First,
    \int_{0}^1\sqrt{[x]}dx=   \int_{0}^a\sqrt{[x]}dx+\int_{1}^2\sqrt{[x]}dx+\cdots +\int_{a-1}^a\sqrt{[x]}dx

    second, note that in the first interval (integral)  [x]=\sqrt{0}
    in the second interval (integral) [x]=\sqrt{1}
    and so on, unitil
    the ath integral  [x]=\sqrt{a-1}, so
    \int_{0}^1\sqrt{[x]}dx=   \int_{0}^a0dx+\int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx
    =   \int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx     =1+2+\cdots+\sqrt{a-1} as desired.
    Hi I have no idea why the first part of your answer is true?
    I don't really even understand what it means?
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  4. #4
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    Re: Please Help with an integration Q

    Quote Originally Posted by James7361539 View Post
    Hi I have no idea why the first part of your answer is true?
    I don't really even understand what it means?
    you are right. I correct the bounds
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  5. #5
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    Re: Please Help with an integration Q

    Quote Originally Posted by Kmath View Post
    you are right. I correct the bounds
    Hey Kmath

    I now understand how to do part 1 but now I am stuck on part 2

    I have attached the graph of y=2^[x] which I think will help.

    I would be grateful if you could please explain how you do part 2

    Thanks
    Attached Thumbnails Attached Thumbnails Please Help with an integration Q-photo-1.jpg  
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  6. #6
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    Re: Please Help with an integration Q

    Ok, again we have
    \int_0^a{2^{[x]}dx}=\int_0^1{2^{[x]}dx}+\int_1^2{2^{[x]}dx}+\cdots+\int_{a-1}^a{2^{[x]}dx}
    as the fisrt part we know the value of {[x]} on each interval, that is
    \int_0^a{2^{[x]}dx}=\int_0^1{2^{0}dx}+\int_1^2{2^{1}dx}+\cdots+ \int_{a-1}^a{2^{a-1}dx}
    =\int_0^1{1dx}+\int_1^2{2dx}+\cdots+\int_{a-1}^a{2^{a-1}dx}
    All the integrants are constants and the lenght of of each interval is 1 we have
    \int_0^a{2^{[x]}dx}=1+2+4+\cdots+{2^{a-1}}
    The later series is finite geometric series and its sum gives you the result
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  7. #7
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    Re: Please Help with an integration Q

    I just love your forum.Thanks for posting it.Web Development it have something that someone comeback again….there is a lot of useful information a person can get from here…I must say,well done.Free Legal AdviceA good forum with great discussion and a good users,which contribute in the forum.
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  8. #8
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    Re: Please Help with an integration Q

    Quote Originally Posted by Kmath View Post
    Ok, again we have
    \int_0^a{2^{[x]}dx}=\int_0^1{2^{[x]}dx}+\int_1^2{2^{[x]}dx}+\cdots+\int_{a-1}^a{2^{[x]}dx}
    as the fisrt part we know the value of {[x]} on each interval, that is
    \int_0^a{2^{[x]}dx}=\int_0^1{2^{0}dx}+\int_1^2{2^{1}dx}+\cdots+ \int_{a-1}^a{2^{a-1}dx}
    =\int_0^1{1dx}+\int_1^2{2dx}+\cdots+\int_{a-1}^a{2^{a-1}dx}
    All the integrants are constants and the lenght of of each interval is 1 we have
    \int_0^a{2^{[x]}dx}=1+2+4+\cdots+{2^{a-1}}
    The later series is finite geometric series and its sum gives you the result
    Thanks for all your help Kmath.
    I now understand how to do this Q
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