Double integrals

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    Hi, I am having problems setting up a double integral. For the first integration sign, the limits should be the range on the x-axis, and the second integral, the limits should be the range of the y-axis?

    for example, if you have two curves intersecting at x=0 and x=1, and you are told to find the area of the enclosed shape, x=0 and x=1 will be the limits for the outside integral. But how do I know the limits on the inside integral?
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    It's not clear why you're trying to set up a double integral. From the information provided, you should be able to calculate the area between the curves using the difference of two single integrals.
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    (Original post by mik1a)
    It's not clear why you're trying to set up a double integral. From the information provided, you should be able to calculate the area between the curves using the difference of two single integrals.
    The question asks me to solve using a double integral.
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    It's not clear to me why you would use this method. Could you share the question in full?
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    (Original post by cushticwarrior)
    The question asks me to solve using a double integral.
    Let f(x) and g(x) be curves that intersect at x=0 and x=1.

    Then the area between them is \left| \int_{0}^{1}f(x)dx-\int_{0}^{1}g(x)dx \right| =\left| \int_{0}^{1}f(x)-g(x)dx\right|
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    (Original post by Cryptokyo)
    Let f(x) and g(x) be curves that intersect at x=0 and x=1.

    Then the area between them is \left| \int_{0}^{1}f(x)dx-\int_{0}^{1}g(x)dx \right| =\left| \int_{0}^{1}f(x)-g(x)dx\right|
    Surely this is just a single integration? I thought it would be closer to int 0 to 1 followed by int x^2 to x^3

    if we take y=x^2 and y=x^3 to be the curves that intersect


    It doesn't let me do the limits properly with the x square/cubed but I hope you can understand what I'm trying to show

    \int^1_0\ \int^x^3_x^2 dydx

    (Original post by mik1a)
    It's not clear to me why you would use this method. Could you share the question in full?
    The curves y=x^2 and y=x^3 intersect ar x=0 and x=1. Use a double integral to calculate the area enclosed by the two curves between x=0 and x=1.
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    Cryptokyo's method is correct (though you must be careful if the curves cross between 0 and 1). It is single integration.

    There is no need to do double integration here. Just calculate int x^2 - int x^3, each from 0 to 1.

    Double integrals are used for functions of two variables.
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    (Original post by mik1a)
    Cryptokyo's method is correct (though you must be careful if the curves cross between 0 and 1). It is single integration.

    There is no need to do double integration here. Just calculate int x^2 - int x^3, each from 0 to 1.

    Double integrals are used for functions of two variables.
    I need to know how to do it since we are studying multiple integration! I need to get the foundations down, like knowing what limits to use on the inside integral are. Do you know how to do it?
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    (Original post by cushticwarrior)
    I need to know how to do it since we are studying multiple integration! I need to get the foundations down, like knowing what limits to use on the inside integral are. Do you know how to do it?
    The question you have quoted doesn't seem to ask you to use double integration, so I'm not sure why you insist you're supposed to use it.

    I mean, you can (trivially) rewrite a standard integral as a double one, but it's just obfuscation - I can't see an actual reason to do it here.
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    (Original post by DFranklin)
    The question you have quoted doesn't seem to ask you to use double integration, so I'm not sure why you insist you're supposed to use it.

    I mean, you can (trivially) rewrite a standard integral as a double one, but it's just obfuscation - I can't see an actual reason to do it here.
    'Use a double integral' it says. It's no good finding the area between two curves using standard integration in an exam if it asks for a double integral. I also thought when we had our first lesson on this last week 'what's the point using double integration for a simple right angled triangle', but that was beside the point.

    It's not as if you have to be in the deep end to start using it.

    If you guys thinks it's that trivial - you don't have to help. I'll try to find it out or ask my teacher next time we meet.
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    To put it in terms you'll understand: it's like asking someone to solve "5x = 10" using trigonometry. It does not make sense.
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    (Original post by cushticwarrior)
    'Use a double integral' it says. It's no good finding the area between two curves using standard integration in an exam if it asks for a double integral. I also thought when we had our first lesson on this last week 'what's the point using double integration for a simple right angled triangle', but that was beside the point.

    It's not as if you have to be in the deep end to start using it.

    If you guys thinks it's that trivial - you don't have to help. I'll try to find it out or ask my teacher next time we meet.
    A double integral implies something of the form \int_{a}^{b}\left(\int_{c}^{d}yd  x\right)dy

    It may mean using two integrals. However, integrating by dx then dy gives you dA (or the area). However you would not use this method until either A2 further maths or uni really.

    Link to double integrals to find area: http://math.harvard.edu/~ytzeng/work...1024_2_sol.pdf

    I get what you mean now.


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    (Original post by cushticwarrior)
    'Use a double integral' it says.
    Yeah, sorry - I got confused about what the actual question was and read the wrong post.

    What I would do is start by finding the double integral for the area under a curve between x= 0 and x = 1.(assume that the curve is always above the x axis). That is, you want the area of the region A = {(x, y): 0<=x<=1, 0 <=y<=f(x)}.

    So your "outside" integral is going to be from x =0 to x =1.

    For a particular value of x, what is the range of values y can take?

    So what limits will your "inner" integral have?

    And finally, you are just wanting the area, so what does your integrand need to be?

    Finally, evaluate the "inner" integral to reduce to a single integral.

    Once you understand how to do this, you can either "fiddle" your original problem to solve it in terms of the simplified scenario above, or you can extend it to directly solve your original problem.
 
 
 
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