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Quick calc 1 question

  1. Jul 16, 2014 #1
    $$\int \sqrt{f(x)}dx$$ is NOT the same as..

    $$\sqrt{\int f(x)dx}$$ right?

    i did an example problem and they turned out not to be equivalent but i just wanted to make certain.




    if they actually aren't equivalent.. why aren't they???
     
  2. jcsd
  3. Jul 16, 2014 #2
    well, better example:

    $$\sqrt{(C\int\sqrt{(a+b)}dx)^2}$$

    this is not the same as..

    $$\sqrt{C^2\int(a+b)dx}$$

    is it?

    (C= constant, a= variable, b=constant (not that it really matters..))
     
    Last edited: Jul 16, 2014
  4. Jul 16, 2014 #3

    PeroK

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    Try ##f(x)=x^2## and you'll see for yourself why they are not the same.
     
  5. Jul 16, 2014 #4
    Remember that the definite integral is a kind of sum (technically a limit of sums), and so one way to understand why things may or may not work with the integral is to compare it to a sum.

    In this case, if you remember that ##\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}## most of the time, and in general ##\sqrt{\sum_1^n a_i}\neq\sum_1^n\sqrt{a_i}##, it's not too hard to see why, most of the time, ##\sqrt{\int_a^bf(x)\ dx}\neq\int_a^b\sqrt{f(x)}\ dx##.
     
  6. Jul 16, 2014 #5

    Thanks this was helpful! :)
     
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