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Math Help - Find area enclosed by means of definite integration

  1. #1
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    Find area enclosed by means of definite integration

    Hello,

    I'm in minor confusion why my answer is wrong as opposed towards the book's answer.

    I'm asked to find the area enclosed by the equations,

     y_{1} = (2x -1)\ and\ y_{2} = (3 + 2x - x^2)

     So\ the\ point\ (s)\ of\ intersection\ implies: (2x - 1) = (3 + 2x - x^2)

    \Rightarrow x = \pm2

    If I compute the area under each equation, and also consider a sign test on both integrals seperately, then:-

     A_{a} = \int ^{2}_{-2} (3 + 2x - x^2)\ dx = \left[(3)(2) + 2^2 - \frac{2^3}{3}\right] - \left[(3)(-2) + 2^2 + \frac{(2)^3}{3}\right]

     A_{a} = \frac{20}{3}\ units

    \Rightarrow A_{a}\ is\ (+ve)

    and,

     A_{b} = \int ^{2}_{-2} (2x -1)\ dx = \left(2^2 -2\right) - \left(2^2 + 2\right) = (-4)\ units

    \Rightarrow A_{b}\ is\ (-ve)

    Now if I consider another sign test,

     Area\,A_{c},\ enclosed\ by\ both\ equations\ = A_{a} \pm A_{b} \Rightarrow A_{c} = (+) + (-)\ or\ (+) - (-) \Rightarrow A_{c}\ is\ (-ve)\ or\ (+ve)

    The total area available under either equation, with limit  x = \pm2 , is  A_a . Then the area  A_{c} , enclosed by  y_{1} and  y_{2} will be less than  A_{a} . So in order for  A_{b} to be subtracted from  A_{a} it must have its sign reversed because it's negative.

    So,

     A_{c} = A_{a} + A{b} = \frac{20}{3} + (-4) = \frac{8}{3}\ units

    However, the book has answer of  \frac{32}{3}\ units

    Therefore,

     A_{c} = \frac{20}{3} - (-4) = \frac{32}{3}\ units

    The enclosed area  A_{c} is greater than  A_a

    I don't understand why my answer is wrong?






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  2. #2
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    Re: Find area enclosed by means of definite integration

    Quote Originally Posted by astartleddeer View Post
    I'm asked to find the area enclosed by the equations,
     y_{1} = (2x -1)\ and\ y_{2} = (3 + 2x - x^2)
     So\ the\ point\ (s)\ of\ intersection\ implies: (2x - 1) = (3 + 2x - x^2)

    The integral should be \int_{ - 2}^2 {\left[ {\left( {3 + 2x - x^2 } \right) - \left( {2x - 1} \right)} \right]dx}

    Graph those together on the same graph and you will see why.
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  3. #3
    Member astartleddeer's Avatar
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    Re: Find area enclosed by means of definite integration

    I've already done that.

    If I can see signs could be an issue I have a habit to treat each integral seperately.

    The majority of the area under y1 is negative.

    I can't see why it's not immiterial if you choose to substract the areas individually or as a lumped integrand?
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    Re: Find area enclosed by means of definite integration

    I've treated the problem as a lumped integrand and I've got the area enclosed as  \frac{32}{3} units.

    But I don't understand how?

    The area requested is part of the area under  y_{2} but this tells me its greater than  A_{a} because for the total area under  y_2 for  x = \pm2 is  \frac{20}{3} .This is what I don't understand.
    Last edited by astartleddeer; December 27th 2012 at 11:59 AM.
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    Re: Find area enclosed by means of definite integration

    Quote Originally Posted by astartleddeer View Post
    I've already done that.

    If I can see signs could be an issue I have a habit to treat each integral seperately.

    The majority of the area under y1 is negative.

    I can't see why it's not immiterial if you choose to substract the areas individually or as a lumped integrand?
    1. If you want to calculate areas you have to use the absolute values of functions because an area below the x-axis is negative and the area above the x-axis is positive.

    2. Normally you can avoid the use of absolute values if you don't integrate "over"(?) a zero. In your case both the straight line and the parabola have zeros in the given interval. That means you have to split the interval. The easiest way is of course to integrate the difference of functions as described by Plato.

    3. In the attached sketch the green area is your first integral, the red area your second.
    Attached Thumbnails Attached Thumbnails Find area enclosed by means of definite integration-enclosarea.png  
    Last edited by earboth; December 27th 2012 at 12:04 PM.
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    Re: Find area enclosed by means of definite integration

    Quote Originally Posted by astartleddeer View Post
    if i can see signs could be an issue i have a habit to treat each integral seperately.

    THAT then is your problem.

    It is one integral of the upper curve minus the lower curve.
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    Re: Find area enclosed by means of definite integration

    I see now

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  8. #8
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    Re: Find area enclosed by means of definite integration

    @ earboth

    What's the computer package you used to build that graph?

    That's another reason why I was going wrong. I didn't realise the area UNDER the x-axis turns itself out onto the outside of the curve.

    Imagine drawing a line, from the parabola, vertically all the way down through x =2 and then stop until you reach the y value where x = -2, that's how I've always pictured the area under any -f(x) parabola.

    This is why I was making the claim on how the segmented area could be possibly bigger than the area under the parabola.

    But it all makes sense now.
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