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

$$1.\int { x } dx=x\int { 1 } dx\\ 2.\int { t } dx=t\int { 1 } dx\\ 3.\int _{ x }^{ x+1 }{ x } dt=x\int _{ x }^{ x+1 }{ 1 } dt$$

Which of the equations are correct?

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 Quote by tahayassen $$1.\int { x } dx=x\int { 1 } dx\\ 2.\int { t } dx=t\int { 1 } dx\\ 3.\int _{ x }^{ x+1 }{ x } dt=x\int _{ x }^{ x+1 }{ 1 } dt$$ Which of the equations are correct?
3 and 2 are both correct.

 Mentor 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.

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

 Quote by Mark44 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.

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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 $$1.\int { x } dx=x\int { 1 } dx$$
No, x is the variable of integration so we cannot take it outside the integral.
The integral on the left is $x^2/2+ C$ and on the right $x(x+ c)= x^2+ cx$.

 $$2.\int { t } dx=t\int { 1 } dx$$
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.

 $$3.\int _{ x }^{ x+1 }{ x } dt=x\int _{ x }^{ x+1 }{ 1 } dt$$
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?

 Thanks for the clear-up. :)