A simple question about integration by substitution

[ScienceNerd36]

Hello all,

We've just begun integration in my maths class and I have a question about a certain aspect of integration by substitution.

Let's say for instance you let u = 2x-1. Then you differentiate it and get du/dx = 2.

My maths teacher said " you can now think of it as multiplying across by 'dx' ".

Which leads to du = 2dx.

My question is, what mathematical operations are you actually committing to achieve this line?

Thanks in advance for any help.

 

[mathman]

Hello all,

We've just begun integration in my maths class and I have a question about a certain aspect of integration by substitution.

Let's say for instance you let u = 2x-1. Then you differentiate it and get du/dx = 2.

My maths teacher said " you can now think of it as multiplying across by 'dx' ".

Which leads to du = 2dx.

My question is, what mathematical operations are you actually committing to achieve this line?

Thanks in advance for any help.

It is a symbolic multiplication, since du and dx are symbols not numbers.

 

[HallsofIvy]

If your integration were, say,
[tex]\int e^{2x- 1}dx[/tex]
you can think in either of two ways:
If u= 2x- 1 then du= 2dx so dx= (1/2)du and the integral becomes
[tex]\int e^{2x- 1}dx= \int e^u((1/2)du)= (1/2)\int e^u du[/tex]

Or if u= 2x- 1 then du 2dx and if we multiply and divide by 2 we get
[tex]\int e^{2x-1}dx= (1/2)(2)\int e^{2x-1}dx= (1/2)\int e^{2x-1}(2dx)= (1/2)\int e^u du[/tex]

Notice that in the first case we are moving the "1/2" outside the integral and in the second case, we are moving the "2" inside the integral. We can do that because the derivative of 2x is a constant. If the integral were
[tex]\int e^{x^2}dx[/tex]
so that taking [itex]u= x^2[/itex] gives [itex]du= 2xdx[/itex], we cannot do that substitution. In fact, that integral is not any "elementary" function.