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Math Help - Order of integration

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Order of integration

    Order of integration has never been my forte. I combines the two things I hate the most...shapes...and inequalities .

    Ok so I don't understand what I am doing wrong here.

    Change the order of integration

    \int_0^{\frac{\pi}{2}}\int_0^{\cos(x)}f(x,y)~dy~dx

    So

    0\leq{x}\leq\frac{\pi}{2}

    and

    0\leq{y}\leq\cos(x)

    \Rightarrow\frac{\pi}{2}\leq\arccos(y)\leq{x}

    \Rightarrow\arccos(y)\leq{x}\leq\frac{\pi}{2}

    and I can see that from the picture that the outer bound should be

    \int_0^{1}

    But could someone give me a quick run through on how to manipulate the inequalities?


    I know the first part is wrong and the second right, but if you wouldn't mind doing both.
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  2. #2
    Moo
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    Hello,

    0 \le x \le \tfrac \pi 2 \implies 0 \le \cos(x) \le 1

    0 \le y \le \cos(x) \le 1 \implies 0 \le y \le 1
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    0 \le x \le \tfrac \pi 2 \implies 0 \le \cos(x) \le 1

    0 \le y \le \cos(x) \le 1 \implies 0 \le y \le 1
    Maybe I am being stupid, maybe it's late or I don't know, sorry to bother you but I have no idea what you did

    0\leq{x}\leq{1}
    \overbrace{\Rightarrow}^{\text{?}}\cos(0)\leq{\cos  (x)}\leq\cos\left(\frac{\pi}{2}\right)

    \Rightarrow{1\leq\cos(x)\leq{0}}

    Or do you have to switch the direction of the inequalities

    and for the second one

    I see how you got it (assuming I got the first) part

    but I don't how your first inequality goes to your second
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  4. #4
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Maybe I am being stupid, maybe it's late or I don't know, sorry to bother you but I have no idea what you did

    0\leq{x}\leq{1}
    \overbrace{\Rightarrow}^{\text{?}}\cos(0)\leq{\cos  (x)}\leq\cos\left(\frac{\pi}{2}\right)

    \Rightarrow{1\leq\cos(x)\leq{0}}

    Or do you have to switch the direction of the inequalities
    Between 0 and \tfrac \pi 2, the cosine function is decreasing

    This is why we reverse the inequality order ! And saying that 1 \le \dots \le 0 is just a nonsense

    and for the second one

    I see how you got it (assuming I got the first) part

    but I don't how your first inequality goes to your second
    Just not taking care of cos(x).
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  5. #5
    Flow Master
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    Quote Originally Posted by Mathstud28 View Post
    Order of integration has never been my forte. I combines the two things I hate the most...shapes...and inequalities .

    Ok so I don't understand what I am doing wrong here.

    Change the order of integration

    \int_0^{\frac{\pi}{2}}\int_{{\color{red}y = }0}^{ {\color{red}y = } \cos(x)}f(x,y)~dy~dx Mr F says: Note the red stuff I've added.

    [snip]
    1. Add the red stuff.

    2. Draw the region defined by the integral terminals.

    It should then be evident that after reversing the order of integration you get:

    \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~dx ~dy.
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    Moo
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    Quote Originally Posted by mr fantastic View Post
    1. Add the red stuff.
    Why ?
    Doesn't the order of dy dx define the order of integration ?

    \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~{\color{red}dx} ~{\color{red}dy}
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Between 0 and \tfrac \pi 2, the cosine function is decreasing

    This is why we reverse the inequality order ! And saying that 1 \le \dots \le 0 is just a nonsense


    Just not taking care of cos(x).
    Oh of course! For then we would have to have that if x_0<x_1

    that f(x_0){\color{red}<}f(x_1)

    Great that was the missing key!
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    1. Add the red stuff.

    2. Draw the region defined by the integral terminals.

    It should then be evident that after reversing the order of integration you get:

    \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~dx ~dy.
    I got that, but I am a really analytical person and graphs mean nothing to me =(
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  9. #9
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    Quote Originally Posted by Moo View Post
    Why ?
    Doesn't the order of dy dx define the order of integration ?

    \int_{y=0}^{y = 1} \int_{x=0}^{x = \cos^{-1} (y)} f(x,y) ~{\color{red}dx} ~{\color{red}dy}
    You'd be amazed at how much easier it is for novices to draw the region defined by the integral terminals when you add the red stuff and hence explicitly state the equations of the curves that bound the region ....... Even the act of doing this forces them to think of the equations of the curves that bound the region.

    An expert obviously doesn't need to do this ......
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  10. #10
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    Quote Originally Posted by Mathstud28 View Post
    I got that, but I am a really analytical person and graphs mean nothing to me =(
    You have my sympathy. I'll try to avoid future replies that expose your handicap. Perhaps some day gene therapy can cure you.
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    You have my sympathy. I'll try to avoid future replies that expose your handicap. Perhaps some day gene therapy can cure you.
    Ahh what a glorious day that will be .

    Here is my other handicap again

    I should have paid more attention in fourth grade

    I have been able to solve all the ones I have tried in the last ten minutes (about 65 or so) but this one...and I am not entirely sure why...

    0<x<1

    x<y<2-x
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