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Math Help - Finding Area by Integration Problem.

  1. #1
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    Finding Area by Integration Problem.

    Determine the area bound by y=-x^2 + x + 1 and y = |x|?

    I'm not sure if my approach is right, can someone please help?

    Firstly i let y=x and y=-x and sub it in y=-x^2 + x + 1 to find the possible intersection points.

    For the substitution of y=x to the curve equation, there are 2 answers for x, one is negative while the other is positive. I choose the positive solution while neglecting the negative solution.

    For the substitution of y=-x to the curve equation, there are also 2 answers for x, one is negative while the other is positive. I choose the negative solution while neglecting the positive solution.

    So for integration to determine the areas, i split into 2 parts. one part is the integration of the negative solution to 0 using y=-x^2 + x + 1 minus y=-x.

    The other part is the integration from 0 to the positive solution using y=-x^2 + x + 1 minus y=x.

    Finally, adding the 2 integrals give me the answer. May i know is this approach correct? Thanks.
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  2. #2
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    Quote Originally Posted by toffeefan View Post
    Determine the area bound by y=-x^2 + x + 1 and y = |x|?

    I'm not sure if my approach is right, can someone please help?

    Firstly i let y=x and y=-x and sub it in y=-x^2 + x + 1 to find the possible intersection points.

    For the substitution of y=x to the curve equation, there are 2 answers for x, one is negative while the other is positive. I choose the positive solution while neglecting the negative solution.

    For the substitution of y=-x to the curve equation, there are also 2 answers for x, one is negative while the other is positive. I choose the negative solution while neglecting the positive solution.

    So for integration to determine the areas, i split into 2 parts. one part is the integration of the negative solution to 0 using y=-x^2 + x + 1 minus y=-x.

    The other part is the integration from 0 to the positive solution using y=-x^2 + x + 1 minus y=x.

    Finally, adding the 2 integrals give me the answer. May i know is this approach correct? Thanks.
    It always helps to draw a picture.


    Anyway, I agree that you will need to split the problem into two parts, one for everything to the left of the y axis, and one for everything to the right of the y axis.

    You also need to notice that in the region bounded by the two functions, -x^2 + x + 1 \geq |x|.


    Left hand intersection:

    |x| = -x when x < 0.

    So -x^2 + x + 1 = -x

    0 = x^2 - 2x - 1

    0 = x^2 - 2x + (-1)^2 - (-1)^2 - 1

    0 = (x - 1)^2 - 2

    2 = (x - 1)^2

    \pm \sqrt{2} = x - 1

    1 \pm \sqrt{2} = x.

    It should be clear that this particular intersection is at x = 1 - \sqrt{2}.


    Right hand intersection:

    |x| = x when x > 0.

    So -x^2 + x + 1 = x

    x^2 = 1

    x = \pm 1.

    It should be clear that this particular intersection is at x = 1.



    So your integrals are:

    \int_{1 - \sqrt{2}}^0 {\left(-x^2 + x + 1\right) - \left(-x\right)\,dx} + \int_0^1 {\left(-x^2 + x + 1\right) - \left(x\right) \,dx}.
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    Thanks! Got it.
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