Fundamental Theorem of Calculus - Variables x and t

[Sir James]

Hello, I'm getting slightly confused by the following so was hoping someone may be able to clear my problem up.

For integrals, if b is the upper limit and a is the lower limit, I will write ∫[b,a].

From the Fundamental Theorem of calculus part 1 we can show that:

if
F(x) = ∫[x,a] f(t)dt
then
F'(x) = f(x)
where F'(x) is the derivative of F(x) with respect to x
I understand the proof so will not detail.

From this, we can deduce part 2, that:

if
G(x) is any anti-derivative of f(x)
then
∫[b,a] f(t)dt = G(b)-G(a) = G(x) (evaluated at b) - G(x) (evaluated at a)
I understand the proof here also.

However, what I don't understand is that part 2 is actually written:

∫[b,a] f(x)dx = G(b)-G(a)

This notation suggests that integrating with respect to the x, being the term after the integrand is acceptable. However, we actually obtain the proof by finding the anti-derivative w.r.t. x when x was the upper limit in the summation.

My question is - what is the step that seems to be missing seeing as in the second law we actually integrate with respect to what was previously denoted as t? I understand that x and t are two different variables but they are obviously closely related - is it a case that t is dependent on x? If we can integrate w.r.t. either x or t surely they are equal and vary on a 1 to 1 ratio?

Thanks for the help,

James

 

[verty]

I'm going to say, there was no variable t, t was just a new name for x. You were always using x. The formula f(t) was just the formula f(x) with x renamed to t, no substitution was made.

Or I suppose one could say, t is a variable with exactly the same rate of change as x. I don't think it matters really, the formula is the same. I much prefer thinking of it as just a name change.

 

[HallsofIvy]

In $\int_a^b f(t)dt$, t is a dummy variable. It does not appear in the final result so it can be changed at will: $\int_a^b f(t)dt= \int_a^b f(x)dx= \int_a^b f(y)dy$, etc.

This is the same as the "dummy index" in a sum: $\sum_{n=0}^5 n^2= \sum_{j=0}^5 j^2$ because they are both equal to $0^2+ 1^2+ 2^2+ 3^2+ 4^2+ 5^2= 0+ 1+ 4+ 9+ 16+ 25= 55$.

 

[paulfr]

Be sure not to lose the forrest for the trees ...

The Fundamental Theorem's meaning is that
Differentiation and Integration are Inverse Functions/Operations

 

[Sir James]

Thanks for the responses Verty, HallsofIvy and Paulfr