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Math Help - Unsplitting Limits of Integration

  1. #1
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    Unsplitting Limits of Integration

    Suppose I was integrating some function f(x) across the interval [a,d].

    Suppose, though, that this function was a piecewise function defined as follows:

    f(x) = \left\{<br />
\begin{array}{c l}<br />
  ug(x) & a \leq x \leq b  \\<br />
  vg(x) & b \leq x \leq c<br />
\end{array}<br />
\right.<br />
    Where u and v are some constants.

    So the integral looks something like this:

     I = \int_a^b u \, g(x) dx + \int_b^c v g(x) \,dx

     = u\int_a^b \, g(x) dx + v\int_b^c g(x) \,dx

    Is there some way to express this as a single integral across the whole interval [a,c]?

    I tried equating my general integral to w\int_a^c g(x) \, dx and solved for w, but the answer would require that I'd have to compute the individiual integrals first before I could combine them. Which makes it rather pointless.

    The reason I ask is that I'm integrating some pretty big, nasty functions, and it would save me a lot of time under exam conditions if I didn't have to write each integral out twice - it seems to me to be a perfect waste of time when they're both the same function being integrated over adjacent intervals!
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  2. #2
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    If G is a primitive for g, then (obviously) \int_a^c f = u\int_a^b g + v \int_b^c g = vG(c) - (v-u)G(b) - uG(a). Is this what you're looking for?
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  3. #3
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    Quote Originally Posted by maddas View Post
    If G is a primitive for g, then (obviously) \int_a^c f = u\int_a^b g + v \int_b^c g = vG(c) - (v-u)G(b) - uG(a). Is this what you're looking for?
    Yes, this is the equation I derived earlier, and it does reduce the tricky integrations I am dealing with down to mere algebra.

    However, what I was really asking was, given a split up integral as

    I = u\int_a^b g + v \int_b^c g

    Is there a way to write this such that, rather than having two integrals of the same thing over two adjacent intervals, you have one integral over one function through the whole interval?
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  4. #4
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    Yes, obviously \int_a^c f......
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  5. #5
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    Quote Originally Posted by maddas View Post
    Yes, obviously \int_a^c f......
    Yes, but f isn't a continuos function on the integral [a,c]. So although you're perfectly at liberty to write the integral in this manner, you can't practically solve the integral in that form - you would simply need to split f up into ug and vg to practically integrate it.

    What I'm asking for is a practically integrable expression.
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