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Math Help - Area of 2 curves -Integration

  1. #1
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    Area of 2 curves -Integration

    I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.

    Find the area of the region trapped between the 2 given curves and the indicated lines.

    f(x) = 3x
    g(x) = x^2
    x = -1, x = -2

    int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)

    Take z(-2) - z(-1) = results

    What I also did was to integrate each function separately and sub in the the range of x and find the difference.

    Apparently, the results I obtain does not tally with the answer sheet that says 31/6

    I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.

    My answer works out to be 3/2.
    Last edited by Darrylcwc; October 8th 2013 at 09:36 AM.
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  2. #2
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    Re: Area of 2 curves -Integration

    First of all, when is 3x \ge x^2 and when is x^2 \ge 3x? To figure this out, find out when they are equal.

    x^2 = 3x when x=0, x=3.

    So, at x=-1, 1 = (-1)^2 > 3(-1) = -3. Hence x^2 \ge 3x on the interval (-\infty,0].

    How about on [0,3]? Well, 1 = (1)^2 < 3(1) = 3, so 3x \ge x^2 on the interval [0,3].

    So, the correct set-up for this would be:
    \int_{-1}^0 (x^2-3x)dx + \int_0^2 (3x-x^2)dx.
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by Darrylcwc View Post
    I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.

    Find the area of the region trapped between the 2 given curves and the indicated lines.

    f(x) = 3x
    g(x) = x^2
    x = -1, x = -2

    int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)

    Take z(-2) - z(-1) = results

    What I also did was to integrate each function separately and sub in the the range of x and find the difference.

    Apparently, the results I obtain does not tally with the answer sheet that says 31/6

    I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.

    My answer works out to be 3/2.
    Graph the functions to visualzse what you are doing

    Area of 2 curves -Integration-untitled2.gif

    Do you see what you can do now: integrate under the parabola and subtruct the area of the triangle
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    Re: Area of 2 curves -Integration

    Where did the negative one came from?
    Equation is equal only at 0 ll -3.
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by Darrylcwc View Post
    Where did the negative one came from?
    Equation is equal only at 0 ll -3.
    You are told to bound the functions between the lines x=-1 and x=2. Finding where the equations are equal only tells you the points where it "may" switch which one is "above" the other.
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by Darrylcwc View Post
    Where did the negative one came from?
    Equation is equal only at 0 ll -3.
    The integration in the orginal post is area from -2 to -1.
    if your question is about the line y = 3x, at x=-2 y=-6, at x=-1 y=3*(-1) = -3
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by votan View Post
    The integration in the orginal post is area from -2 to -1.
    if your question is about the line y = 3x, at x=-2 y=-6, at x=-1 y=3*(-1) = -3
    Oh, oops! I thought it was -1 and 2. Good catch! So then the integral should be

    \int_{-2}^{-1}(x^2-3x)dx
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  8. #8
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    Re: Area of 2 curves -Integration

    It just doesn't make sense.
    Since x = 0 and x =3 is where the entire curve intersect, then why can't I use x = [0,3] as the range?
    graph y=x^2 && y=3x - Wolfram|Alpha=
    It just doesn't make sense mathematically to cover the domain outside of 0 and 3 because the area is not enclosed.

    I'm solving this, otherwise I'm not sleeping.
    Last edited by Darrylcwc; October 8th 2013 at 08:31 PM.
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  9. #9
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by Darrylcwc View Post
    It just doesn't make sense.
    Since x = 0 and x =3 is where the entire curve intersect, then why can't I use x = [0,3] as the range?

    I'm solving this, otherwise I'm not sleeping.
    Check out the graph of the functions (link). You are asked to find the region bounded by four functions. y=x^2, y=3x, x=-1, x=-2. Where does the line x=-1 intersect the graph y=x^2? It intersects it at x=-1. Similarly for y=3x. And x=-2 intersects both functions at x=-2. Figuring out where the two graphs intersect can help you find "a" bounded region. It is not the only one. There are several bounded regions. Since you need to use all four functions as bounds, the only bounded region with x=-2 and x=-1 as bounds has area as I gave in the previous post.
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  10. #10
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    Re: Area of 2 curves -Integration

    Hang on. Isn't x the domain? Why is it a function?
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  11. #11
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by Darrylcwc View Post
    Hang on. Isn't x the domain? Why is it a function?
    Because you said it was:

    Quote Originally Posted by Darrylcwc View Post
    Find the area of the region trapped between the 2 given curves and the indicated lines.

    f(x) = 3x
    g(x) = x^2
    x = -1, x = -2
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  12. #12
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    Re: Area of 2 curves -Integration

    In this case, shouldn't the set up thus be

    [-1,-2]int (x^2-3x) + [0,3]int (x^2 - 3x) ?
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by Darrylcwc View Post
    In this case, shouldn't the set up thus be

    [-1,-2]int (x^2-3x) + [0,3]int (x^2 - 3x) ?
    notice athe -2 < -1, you could integrate from -1 to -2 if you place a negative sign infront of the integral.

    I couldn't post a new reply, I append this post by the image.

    Area of 2 curves -Integration-untitled2.gif
    Last edited by votan; October 8th 2013 at 10:09 PM.
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  14. #14
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    Re: Area of 2 curves -Integration

    I understand that. This set up I gave should be correct. But it still doesn't tally.
    Answer sheet states 31/6.

    Is something wrong? It's can't be the case. This is supposed to be a straight forward and simple abstract question.
    Last edited by Darrylcwc; October 8th 2013 at 09:26 PM.
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  15. #15
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    Re: Area of 2 curves -Integration

    Quote Originally Posted by SlipEternal View Post
    First of all, when is 3x \ge x^2 and when is x^2 \ge 3x? To figure this out, find out when they are equal.

    x^2 = 3x when x=0, x=3.

    So, at x=-1, 1 = (-1)^2 > 3(-1) = -3. Hence x^2 \ge 3x on the interval (-\infty,0].

    How about on [0,3]? Well, 1 = (1)^2 < 3(1) = 3, so 3x \ge x^2 on the interval [0,3].

    So, the correct set-up for this would be:
    \int_{-1}^0 (x^2-3x)dx + \int_0^2 (3x-x^2)dx.
    I contrast my set up with yours and I don't understand why the second set up ranges from 0,2. The point of intersection is at 0,3.
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