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  1. #1 integration 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    We use integration to solve the area under the curve right? when two line, or curves cut each other forming a shaded region in a Cartesian plane, and you just subtract the upper curve by the lower curve and integrate respect to whatever variable given with the two given x-axis points.
    but my question here is, how if your region of area is surrounded by 3 curves or lines, with three sets of coordinates??

    can anyone help me, though my question is alittle vague in a way, any help would be great. thank you


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  3. #2 Re: integration 
    . DrRocket's Avatar
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    Quote Originally Posted by Heinsbergrelatz
    We use integration to solve the area under the curve right? when two line, or curves cut each other forming a shaded region in a Cartesian plane, and you just subtract the upper curve by the lower curve and integrate respect to whatever variable given with the two given x-axis points.
    but my question here is, how if your region of area is surrounded by 3 curves or lines, with three sets of coordinates??

    can anyone help me, though my question is alittle vague in a way, any help would be great. thank you
    You can either break the region up into a disjoint union of areas bounded by two curves (plus perhaps some verticals) or you can integrate over a region using multivariable calculus.


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  4. #3  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    thank you for the reply, but how would you apply multivariable calculus here?

    let me gIve you the exact question, can you try and solve it? ( im not trying to be lazy, i have attempted it, and my answer isnt exactly correct)
    the two line and one curve is given by,

    and you will probably see the region, and the three points of intersections are,
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  5. #4  
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    The easiest way would be to split up your integration bounds. Evaluate the integral of x^2 from 1 to 2 minus the integral of y=1 from 1 to 2. Then add it with the integral of 10 - 3x from 2 to 3 minus the integral of 1 from 2 to 3.
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  6. #5  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    how about you get the area of the trapezium, and subtract the integral respect to y
    then you would get the same answer, as the one you stated
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