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Integration: what variables can you move outside of the integrand?

 
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Jan24-13, 04:29 PM   #1
 

Integration: what variables can you move outside of the integrand?


[tex]1.\int { x } dx=x\int { 1 } dx\\ 2.\int { t } dx=t\int { 1 } dx\\ 3.\int _{ x }^{ x+1 }{ x } dt=x\int _{ x }^{ x+1 }{ 1 } dt[/tex]

Which of the equations are correct?
 
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Jan24-13, 05:09 PM   #2
 
Quote by tahayassen View Post
[tex]1.\int { x } dx=x\int { 1 } dx\\ 2.\int { t } dx=t\int { 1 } dx\\ 3.\int _{ x }^{ x+1 }{ x } dt=x\int _{ x }^{ x+1 }{ 1 } dt[/tex]

Which of the equations are correct?
3 and 2 are both correct.
 
Jan24-13, 07:00 PM   #3
 
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And 1 is incorrect. The following is a property of integrals:
##\int k~f(x)~dx = k\int f(x)~dx##, for k a constant, but there is no property that says you can move a variable across the integral sign.
 
Jan24-13, 07:53 PM   #4
 

Integration: what variables can you move outside of the integrand?


Quote by Mark44 View Post
And 1 is incorrect. The following is a property of integrals:
##\int k~f(x)~dx = k\int f(x)~dx##, for k a constant, but there is no property that says you can move a variable across the integral sign.
Integral xdx is certainly not the same as x times integral dx.
 
Jan25-13, 09:16 AM   #5
 
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You can move constants (and so variables that are independent of the variable of integration and so are treated like constants in the integration) outside the integral.

Quote by tahayassen View Post
[tex]1.\int { x } dx=x\int { 1 } dx[/tex]
No, x is the variable of integration so we cannot take it outside the integral.
The integral on the left is [itex]x^2/2+ C[/itex] and on the right [itex]x(x+ c)= x^2+ cx[/itex].

[tex] 2.\int { t } dx=t\int { 1 } dx[/tex]
If we know that t is independent of x, then both integrals are "tx+ C". If t is a function of x then the first is still "tx+ C" but the other depends upon exactly what function of x t is.

[tex] 3.\int _{ x }^{ x+1 }{ x } dt=x\int _{ x }^{ x+1 }{ 1 } dt[/tex]
If x is independent of the variable of integration, t, both of those are the same and are equal to x(x+1- x)= x. If x is a function of t, then the left depends upon exactly what function of t x is while the right is still x.

Which of the equations are correct?
 
Jan26-13, 11:52 AM   #6
 
Thanks for the clear-up. :)
 
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