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Math Help - Find the area

  1. #1
    Junior Member R3ap3r's Avatar
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    Find the area

    Find the exact area of the region between the curve y = 4x^3 - 2x^2 - 12x and the x=axis, over the interval [-1, 2]

    My Work:

    \int_{-1}^2\;(4x^3 - 2x^2 - 12x)\;dx = x^4 - \frac{2}{3}\; x^3 - 6x^2

    Let\; x = 2\;\;\;\;\; 2^4 - \frac{2}{3}\; (2)^3 - 6(2)^2

    Let\; x = -1\;\;\;\;\; (-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2

    \left[2^4 - \frac{2}{3}\;(2)^3 - 6(2)^2\right] - \left[(-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2\right]

    = 3.75\; units^2

    Hows my answer look here?
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  2. #2
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    Quote Originally Posted by R3ap3r View Post
    Find the exact area of the region between the curve y = 4x^3 - 2x^2 - 12x and the x=axis, over the interval [-1, 2]

    My Work:

    \int_{-1}^2\;(4x^3 - 2x^2 - 12x)\;dx = x^4 - \frac{2}{3}\; x^3 - 6x^2

    Let\; x = 2\;\;\;\;\; 2^4 - \frac{2}{3}\; (2)^3 - 6(2)^2

    Let\; x = -1\;\;\;\;\; (-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2

    \left[2^4 - \frac{2}{3}\;(2)^3 - 6(2)^2\right] - \left[(-1)^4 - \frac{2}{3}\;(-1)^3 - 6(-1)^2\right]

    = 3.75\; units^2

    Hows my answer look here?
    Looks good again. For you, the power rule is obviously not a problem. However, you must be careful that you are answering the question correctly. If there is a root of the equation between -1 and 2, you have a problem.
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  3. #3
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    Yeah, the root x = 0 is going to cause you problems. Let f(x) = 4x^3 - 2x^2 - 12x. Then you want to integrate \int_{-1}^{0} f(x) and \int_{0}^{2} f(x) separately, and add the absolute values of the answers.
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  4. #4
    Junior Member R3ap3r's Avatar
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    so you saying mines wrong then?
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by R3ap3r View Post
    so you saying mines wrong then?
    Yes look at the graphs below

    Find the area-int1.jpg

    Find the area-int2.jpg

    we need to break it up into two integrals

    \int_{-1}^{0}f(x)dx-\int_{0}^{2}f(x)dx

    The minus sign is because the area is below the x axis.

    I hope this clears it up.
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  6. #6
    Junior Member R3ap3r's Avatar
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    I c. how do you show your working for that exactly?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    State that the curve is less than 0 on [-1,0] and greater than 0 on [0,2] then state that to get the area of a curve when it is below the x-axis you must take the opposite of it...evaluate the integrals..add and poof! you have the answer
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  8. #8
    Junior Member R3ap3r's Avatar
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    what I meant to say was what did you get as your answer for it emptyset? does it differ a lot from my own?
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by R3ap3r View Post
    what I meant to say was what did you get as your answer for it emptyset? does it differ a lot from my own?
    Taking \int_{-1}^{0}f(x)dx-\int_0^{2}f(x)dx=\frac{13}{3}+\frac{40}{3}=\frac{5  3}{3}square units
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  10. #10
    Junior Member R3ap3r's Avatar
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    so where exactly did i go wrong in my first post?
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by R3ap3r View Post
    so where exactly did i go wrong in my first post?
    Technically the area between any curve and the x-axis is \int|f(x)|dx

    So when the curve changed signs you had to break the integral at that piont and evaluate them seperately based on whether due to the || if it was positive or negative
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