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Math Help - Integrating "lumps" instead of "pieces"

  1. #1
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    Question Integrating "lumps" instead of "pieces"

    Hello everyone,

    In one of my math courses we've just finished covering double integrals and our professor had mentioned that when evaluating double integrals, the work of evaluating certain double integrals will greatly simplify if you choose to integrate in "lumps" as oppose to "bits and pieces".

    I'll illustate what I mean by lumps and pieces.

    If we are given the following double integral,

    <br />
\int^{1}_{0} \int^{x-1}_{0} (4x + y)^{3}dydx

    This is an integral which would be much easier if we integrated it in lumps giving us,

    \frac{1}{4}\int_{0}^{1} [(4x+y)^{4}]^{x-1}_{0}dx

    etc...

    However it's not always possible to integrate in lumps. For example if we had,


    <br />
\int^{1}_{0} \int^{x-1}_{0} (4x + y^{2})^{3}dydx

    We could not preform an integration of the lump in this case, we must expand until our integral is in mutliple pieces and integrate the pieces individually.

    Now what I'm trying to think of is how I can consistently identify and determine whether an given integral can integrated in a lump, or whether it must be done in pieces.


    Is there a simple way of doing this? What are the trouble cases that we cannot integrate in a lump?

    Thanks again!
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  2. #2
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    Staten Island, NY
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    If I'm understanding correctly, then your notion of "integrating in a lump" is the same thing as applying u-substitution. Your notion of "integrating in pieces" means u-substitution can't be applied, so you need to do some algebraic manipulation.

    In general, there is no absolute way to know that u-substitution (lumps) can be used. Sometimes you just need to be clever. But there is a general guideline:

    Remember that u-substitution for integration is "undoing" a chain rule for differentiation. Thus you want to look for the inner part of a composition whose derivative lies outside the composition (up to a constant).

    In your first example, the derivative of 4x + y is just 1.

    In your second example, the derivative of 4x + y^2 is 2y, which is not in the problem.

    If the integrand was (4x+y^2)^3 (2y), then a u-substitution (lump) would work.

    Hope this helps!
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