wispy ruin
My professor uses this definition of the definite integral to segue into the fundamental theorem of calculus, however i'm very confused about the use of t here.
I was told it's a "placeholder" but if A(x) differentiates to equal f(x), then does f(t) = f(x)?
cerulean prawn
My professor uses this definition of the definite integral to segue into the fun...
t is a dummy variable for integration: it disappears after you insert the limits. You can replace it by any letter, as long it is not the one used in limits.
wispy ruin
but what does is represent? what exactly is a dummy variable
cerulean prawn
but what does is represent? what exactly is a dummy variable
Well, what happens when you find the integral of f(t)? You find the antiderivative F(t).
Then, using the fundamental theorem of calculus, you substitute the limits. Thus, the integral is equal to F(x) - F(a).
So, the variable t is just for convenience of finding the antiderivative.
mighty estuary
This is a thing we do all the time in calculus; there are a lot of functions that are defined as the integral from some constant to x of some function.
ln is one, in fact. ln(x) := int(1, x) 1/t dt.
wispy ruin
how is it more convenient to find the antiderivative when the variable is t
mighty estuary
how is it more convenient to find the antiderivative when the variable is t
It's not about convenience.
wispy ruin
we're not all that far into ftc but i just want to know what exactly t represents
mighty estuary
It's just that one of your bounds of integration is x, so your variable of integration can't be x.
cerulean prawn
how is it more convenient to find the antiderivative when the variable is t
So you don't confuse it with x that is present in the limits.
If the limits were just constants or variables other than x, then you could've written x.
wispy ruin
yeah i mean but graphically speaking, what does f(t)dt represent
mighty estuary
yeah i mean but graphically speaking, what does f(t)dt represent
...okay, switch the x and the t.
If that's what's confusing you, switch the x and the t. Look at it as int(a, t) f(x) dx.
Bearing in mind that a is constant and t is variable.
cerulean prawn
yeah i mean but graphically speaking, what does f(t)dt represent
Same as usual: a rectangle with height f(t) and width dt.
wispy ruin
would it be correct to assume that f(t) is like the curve that is being measured?
cerulean prawn
f(t) is the function we're integrating.
wispy ruin
so f(t) could be like t^2 or something, and all x is doing is just setting the upper limit?
cerulean prawn
Yes.
wispy ruin
so what's the deal with this then? if you differentiate A(x), shouldn't you just get f(t)?
cerulean prawn
No.
By the fundamental theorem of calculus. the integral is F(x) - F(a). So:
A'(x) = (F(x) - F(a)) = F'(x) = f(x)
mighty estuary
so what's the deal with this then? if you differentiate A(x), shouldn't you just...
...no, because you're not differentiating A(t), you're differentiating A(x).
mighty estuary
And you're differentiating with respect to x.
cerulean prawn
Well, they are not equal, but they represent the same thing.
mighty estuary
does f(x) = f(t)?
If x = t.
wispy ruin
so f(t) is just the function being integrated, A(x) is the area below it, and f(x) is the derivative of that area function which happens to represent the same thing as f(t)
cerulean prawn
Yes.
mighty estuary
so f(t) is just the function being integrated, A(x) is the area below it, and f(...
Well. A(x) - A(a) is the area below it from a to x.
But a and therefore A(a) is a constant.
cerulean prawn
Well. `A(x) - A(a)` is the area below it from a to x.
No, I believe that just A(x) is the area, considering the definition of A(x) given.
A(x) = F(x) - F(a) for any antiderivative F(x), though.
wispy ruin
which is equal to the int(a, x) f(t)dt?
cerulean prawn
A(x)? Yes.