Question

Why is it that you can't combine the numerator and denominator into one integral and then cancel before integrating, like in the following example?
\[\bar{x} = \frac{\int\limits \sin^2(x)dx}{\int\limits \sin(x)dx }\]

Answer 1

yor example seems to be:\[\bar{x} = \frac{\int \sin^2(x) \; dx}{\sin(x)}\]

but
\[\int \sin^2(x) \; dx \ne \sin^2 x\]

so you are cancelling a ghost

Answer 2

Sorry, I had a typo in my LaTex equation. I meant the following:\[\bar{x} = \frac{\int\limits \sin^2(x)dx}{\int\limits \sin(x)dx }\]

Answer 3

i think you want to say that \(\bar{x} = \frac{\int \sin^2(x)dx}{\int \sin(x)dx} \) equals \( \int \sin(x) \; dx\)??

but you can disprove that easily in this specific case by doing the actual work.

moreover you can look at it generally in any number of ways. for example is there any algebraic reason why \(\dfrac{\sum_1^\infty \sin^2 x_i \Delta x_i}{\sum_i^\infty \sin x_i \Delta x_i}\) should equal \(\sum_1^\infty \sin x_i \Delta x_i\)???

and you might wish to differentiate \(\bar{x} = \frac{\int \sin^2(x)dx}{\int \sin(x)dx}\) too, just to see where it leads

hope that was in some way helpful....:p

Answer 4

here is a hint :-
\(\Large \dfrac{\int g(x) dx}{\int f(x) dx}\neq \int{\dfrac{ g(x) }{ f(x) } dx} \)
you just need one counter example
lets make g(x)=1 and f(x)=x
\(\Large \dfrac{\int 1~ dx}{\int x~ dx}=\dfrac{x+c_1}{0.5x^2+c_2}\\\Large\neq \int{\dfrac{ 1 }{ x } dx} =\log x+c\)

Answer 5

and also there is something you need to know

\(\Large \int (f(x))^2 ~dx \neq (\int (f(x) ~dx)^2 \)
in general
\(\Large \int (f(x))^n ~dx \neq (\int (f(x) ~dx)^n \)

Answer 6

another property that also you could make use of

\(\Large \int f(x).g(x) ~dx \neq \int f(x) ~dx.\int g(x) ~dx \)

Answer 7

thank you Ikkles, so much better.

i am no good at maths so i am struggling to express myself, but i think Doughnut wants:
\(\dfrac{F(x)}{G(x)} = \int \dfrac{f(x)}{g(x)} \; dx\)

in my little world, for that you need :

\(\dfrac{fG - Fg}{G^2} = \dfrac{f}{g}\)

sure it exists, but you gotta find it!!!

Answer 8

well its a property just like saying subtraction on N is not commutative hmm as
\(a-b\neq b-a\) but there is cases when a=b blah bllah, that does not mean the statement is correct in general.
another example \(\dfrac{a+b}{a+c}\neq a(\dfrac{b}{c})\) which sounds like your case, but we might find some a,b,c that satisfy it (and we might not).