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Math Help - Area Integral

  1. #1
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    Area Integral

    I need to find the area of the region between the parabpla y^2 = 16x an the line 4x-3y=4.

    I understand that i need to get them to both equal y but when i do that i seem to confuse myself and get really odd values.
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  2. #2
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    Quote Originally Posted by menco View Post
    I understand that i need to get them to both equal y but when i do that i seem to confuse myself and get really odd values.
    Certainly we don't obtain rational solutions ( it appears \sqrt{13} ) but, what is the problem?


    Fernando Revilla
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    In my text book it shows that i should have the curves equal to y. So that I can find the x intercepts and work with the integral from there. However I can seem to figure out a solution for y^2=16x and 4x-3y=4. When I tried i got y=4 sqrt x and y=-(4(-x-1)/3) but i dont think thats right.
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  4. #4
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    Quote Originally Posted by menco View Post
    In my text book it shows that i should have the curves equal to y. So that I can find the x intercepts and work with the integral from there. However I can seem to figure out a solution for y^2=16x and 4x-3y=4. When I tried i got y=4 sqrt x and y=-(4(-x-1)/3) but i dont think thats right.
    If you're studying calculus you're expected to know how to solve simultaneous equations.

    I suggest finding the area between x = y^2/16 and x = (4 + 3y)/4, in which case you need the y-coordinates of the intersection points of the two curves.

    Equate: y^2/16 = (4 + 3y)/4. Solve for y.
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  5. #5
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    I got a result of 13.2111, -1.2111 for y co ordinates. Using this i integrated f(x)((4-3y)/4) - g(x) (y^2/16) using the y values on the integral. I got an overall result of 3.004 units^2.

    Am i close?
    Last edited by menco; January 27th 2011 at 10:26 PM.
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    Quote Originally Posted by menco View Post
    I got a result of 13.2111, -1.2111 for y co ordinates. Using this i integrated f(x)((4-3y)/4) - g(x) (y^2/16) using the y values on the integral. I got an overall result of 3.004 units^2.

    Am i close?
    No.

    Your values of y are correct (to 4dp - but are you meant to use exact values?)

    Using your values of y as the integral terminals, I don't get your answer (but since you only give a final answer, it's impossible to know whether your error is in the calculus, algebra or arithmetic).
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  7. #7
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    How do people put the working out as an image? If you know how they do this I will post up how I got to my answer
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    Quote Originally Posted by menco View Post
    How do people put the working out as an image? If you know how they do this I will post up how I got to my answer
    Click on the appropriate link in my signature.
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  9. #9
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    \int_{-1.2111}^{13.2111}\left[(\frac{4+3y}{4}\right)-(\frac{y^2}{16})] dx

    [(\frac{4y+(3y^2/2)}{4y})-(\frac{(y^3/3)}{16y})]_{-1.2111}^{13.2111}

    (\frac{4(13.2111)+(3(13.2111)^2/2)}{4(13.2111)})-(\frac{(13.2111)^3/3)}{16(13.2111)})-(\frac{4(-1.2111)+(3(-1.2111)^2/2)}{4(-1.2111)})-(\frac{(-1.2111)^3/3)}{16(-1.2111)})

    I had another go the way i was intending to do it in the first place. and got a result of 1.742.

    The last way I tried to combine the 2 fractions to make it a bit easier it looked like this
    \int_{-1.2111}^{13.2111}\left(\frac{4+3y-y^2}{4-16}\right) dx

    But i think i got that wrong
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  10. #10
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    Quote Originally Posted by menco View Post
    \int_{-1.2111}^{13.2111}\left[(\frac{4+3y}{4}\right)-(\frac{y^2}{16})] dx

    [(\frac{4y+(3y^2/2)}{4y})-(\frac{(y^3/3)}{16y})]_{-1.2111}^{13.2111}

    (\frac{4(13.2111)+(3(13.2111)^2/2)}{4(13.2111)})-(\frac{(13.2111)^3/3)}{16(13.2111)})-(\frac{4(-1.2111)+(3(-1.2111)^2/2)}{4(-1.2111)})-(\frac{(-1.2111)^3/3)}{16(-1.2111)})

    I had another go the way i was intending to do it in the first place. and got a result of 1.742.

    The last way I tried to combine the 2 fractions to make it a bit easier it looked like this
    \int_{-1.2111}^{13.2111}\left(\frac{4+3y-y^2}{4-16}\right) dx

    But i think i got that wrong
    You need to learn how to integrate.

    \displaystyle \int \frac{4 + 3y}{4} \, dy \neq \frac{4y + \frac{3}{2} y^2}{4y}.

    \displaystyle \int \frac{y^2}{16} \, dy \neq \frac{\frac{1}{3} y^3}{16y}.
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  11. #11
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    I am definatly struggling with integration. Can you suggest any good site i can learn from? I find my textbook very hard to follow
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  12. #12
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    Quote Originally Posted by menco View Post
    I am definatly struggling with integration. Can you suggest any good site i can learn from? I find my textbook very hard to follow
    You must have learnt the rule \displaystyle \int a y^n \, dy = \frac{a}{n+1} y^{n+1} (ignoring the arbitrary constant of integration), n \neq -1.

    Both of your integrals are of this type. There is no point doing questions requiring the application of integration if you are struggling with things like this.
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  13. #13
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    Ok so i think i may have figured this out

    (\frac{1}{8}y(3y+8)-(\frac{1}{48}y^3)_{-1.211102551}^{13.21110255}

    And i get a final result of 28.80033927
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  14. #14
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    Quote Originally Posted by menco View Post
    Ok so i think i may have figured this out

    (\frac{1}{8}y(3y+8)-(\frac{1}{48}y^3)_{-1.211102551}^{13.21110255}

    And i get a final result of 28.80033927
    No. The first integral is wrong. Do you realise that \displaystyle \frac{4 + 3y}{4} = 1 + \frac{3}{4} y and so the integral is .....

    You are strongly advised to review basic integration techniques (and the pre-requisite algebra) before attempting questions involving applications of integration.
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