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

[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?

 

[STEMucator]

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

Which of the equations are correct?

3 and 2 are both correct.

 

[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.

 

[STEMucator]

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]

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.

$$1.\int { x } dx=x\int { 1 } dx$$

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?

 

[tahayassen]

Thanks for the clear-up. :)