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


by tahayassen
Tags: integrand, integration, variables
tahayassen
tahayassen is offline
#1
Jan24-13, 04:29 PM
P: 273
[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?
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Zondrina
Zondrina is offline
#2
Jan24-13, 05:09 PM
P: 1,335
Quote 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.
Mark44
Mark44 is online now
#3
Jan24-13, 07:00 PM
Mentor
P: 21,030
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.

Zondrina
Zondrina is offline
#4
Jan24-13, 07:53 PM
P: 1,335

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


Quote 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.
HallsofIvy
HallsofIvy is offline
#5
Jan25-13, 09:16 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886
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 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?
tahayassen
tahayassen is offline
#6
Jan26-13, 11:52 AM
P: 273
Thanks for the clear-up. :)


Register to reply

Related Discussions
iterative integration in several variables Calculus 0
integration by parts of a dot product scalar integrand Advanced Physics Homework 7
change of variables in integrand Calculus 0
Integration of many variables Calculus & Beyond Homework 2
Appropriate Change of Variables for integration Calculus 6