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Math Help - Integral Problem

  1. #1
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    Integral Problem

    Hi

    can anyone help with this one?


    (integral) dx/x^2 +2 dx

    where the limits are b= srt2 and a=0


    MathCad gives answer of 1/8pi2^0.5 but not sure how to get there by hand
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  2. #2
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    Quote Originally Posted by wannberocketscientist View Post
    Hi

    can anyone help with this one?


    (integral) dx/x^2 +2 dx

    where the limits are b= srt2 and a=0


    MathCad gives answer of 1/8pi2^0.5 but not sure how to get there by hand
    \int_0^{\sqrt{2}} \frac{dx}{x^2 + 2}

    Let y = \frac{x}{\sqrt{2}} \implies dy = \frac{dx}{\sqrt{2}} thus
    \int_0^{\sqrt{2}} \frac{dx}{x^2 + 2} = \int_0^1 \frac{\sqrt{2}~dy}{2y^2 + 2}

    = \frac{\sqrt{2}}{2} \int_0^1 \frac{dy}{y^2 + 1}

    Does this form give you any ideas about how to proceed?

    -Dan
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  3. #3
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    not sure do i now just apply :

    f(x) = F(b) - F(a)?
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  4. #4
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    Yes but you first have to get rid of that integral. Because you can't simply fill in an x value into that.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wannberocketscientist View Post
    not sure do i now just apply :

    f(x) = F(b) - F(a)?
    In this case it would be \int_0^1 f(y)~dy = F(1) - F(0), yes. Do you know what the function F is?

    -Dan
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  6. #6
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    Cool

    got it:

    2/srt2 * arctan(1) - arctan(0)= 0.5553603673

    MathCad Solution :

    1/8*pi*2^0.5 = 0.553603673

    So now that these match I need to express the answer in terms of pi - I am assuming that this means I have to turn the numerical value in terms of pi (e.g. pi/2)???

    However this seems like an ugly solution so to speak as it would give:

    5.66pi

    doesnt sound right
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  7. #7
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    Arctan(x)\Big|^1_0 = \frac{\pi}{4} - 0

    Remember that tan(45 deg) = 1, so the tan of pi/4 is 1. So the Arctan of 1 is pi/4.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wannberocketscientist View Post
    got it:

    2/srt2 * arctan(1) - arctan(0)= 0.5553603673
    And note that the original expression from my solution used \frac{\sqrt{2}}{2}, not \frac{2}{\sqrt{2}}

    -Dan
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  9. #9
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    that was a typo sorry!
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  10. #10
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    Dan,

    In your first post here you change the expression to put srt2 and 2 into the fraction. Could you explain where the 2 comes from that is placed in front of the x2?

    Thanks.
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  11. #11
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    Well he did y = \frac{x}{\sqrt{2}}. So basically he took that and solved for x to get x = \sqrt{2}y. Then he needed to square that because the question has x^2.

    I would done it in this order though:

    \int \frac{dx}{x^2 + 2} = \int \frac{dx}{2((\tfrac{x}{\sqrt{2}})^2 + 1)}

    Then you can see that the substitution of y= x/sqrt(2) is quite simple.
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wannberocketscientist View Post
    Dan,

    In your first post here you change the expression to put srt2 and 2 into the fraction. Could you explain where the 2 comes from that is placed in front of the x2?

    Thanks.
    To be honest, when I did the problem the first time I put x^2 = 2y^2 because that's what I saw to be useful. But I thought that putting the substitution in this form would be more confusing than the one that I chose. Perhaps not.

    -Dan
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