Quick calc 1 question

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  • #1
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$$\int \sqrt{f(x)}dx$$ 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???
 

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  • #2
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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..))
 
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  • #3
PeroK
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$$\int \sqrt{f(x)}dx$$ 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???

Try ##f(x)=x^2## and you'll see for yourself why they are not the same.
 
  • #4
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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##.
 
  • #5
466
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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##.


Thanks this was helpful! :)
 

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