Question about Indefinite integrals

[Flumpster]

Homework Statement

I hope this is in the right forum, because this is a question on theory and not related to a specific problem.

I was reading onlne about the Fundamental Theorem of Calculus. On one site the author wrote:

F(x) = \int_{0}^{x} f(t) dt

Later, he wrote:

\int_{a}^{b} f(x) dx = \int_{0}^{b} f(x) dx - \int_{0}^{a} f(x) dx = F(b) - F(a)



However, I've been taught that

\int_{a}^{b} f(x) dx

equals the indefinite integral of f(x) evaluated at b minus the indefinite integral of f(x) evaluated at a.

This leads me to ask:

Is the indefinite integral of a function, let's call it G(x), the same as the definite integral \int_{0}^{x} f(t) dt ?

I know this is a really basic question.

****
Edited to fix what Mark44 pointed out

 

[Mark44]

Homework Statement

I hope this is in the right forum, because this is a question on theory and not related to a specific problem.

I was reading onlne about the Fundamental Theorem of Calculus. On one site the author wrote:

F(x) = \int_{0}^{1} f(x) dx

I don't think you have copied this correctly. What you have above is a number, not a function. The value of the integral doesn't depend on x at all.

It's more likely that the author wrote something like
$$F(x) = \int_{0}^{x} f(t) dt $$

This is now a function. Notice that the variable x appears as one of the limits of integration. The variable t (and dt) are called dummy variables. We could just as well have used r and dr or whatever, as long as we don't use x and dx.

Later, he wrote:

\int_{a}^{b} f(x) dx = \int_{0}^{b} f(x) dx - \int_{0}^{a} f(x) dx = F(b) - F(a)

You can write
$$ \int_{a}^{b} f(x) dx $$
as
$$ \int_{a}^{0} f(x) dx + \int_{0}^{b} f(x) dx $$
Switch the limits of integration on the first integral to get
$$ -\int_{0}^{a} f(x) dx + \int_{0}^{b} f(x) dx $$
This is the same as -F(a) + F(b), using F as defined in the revised definition of F that I showed above.

However, I've been taught that

\int_{a}^{b} f(x) dx

equals the indefinite integral of f(x) evaluated at b minus the indefinite integral of f(x) evaluated at a.

Better terminology would be "antiderivative of f" instead of "indefinite integral of f".

This leads me to ask:

Is the indefinite integral of a function, let's call it G(x), the same as the definite integral \int_{0}^{x} f(x) dx ?

Let me set things up a little better.
Here's an indefinite integral:
$$\int f(x) dx$$

Let F(x) be defined as:
$$ F(x) = \int_0^x f(t)dt$$

Then, F'(x) = f(x) (by the Fund. Thm. of Calc), so F is an antiderivative of f.

By the other part of the FTC, you can evaluate a definite integral by using an antiderivative.

$\int_a^b f(t)dt = \int_0^b f(t)dt - \int_0^a f(t)dt = F(a) - F(b)$

 

[Flumpster]

Mark44 -

You're right, I miss-copied. When I have access to a computer I'll try to explain what I meant.
Thanks :)

 

[Flumpster]

Let me set things up a little better.
Here's an indefinite integral:
$$\int f(x) dx$$

Let F(x) be defined as:
$$ F(x) = \int_0^x f(t)dt$$

Then, F'(x) = f(x) (by the Fund. Thm. of Calc), so F is an antiderivative of f.

By the other part of the FTC, you can evaluate a definite integral by using an antiderivative.

$\int_a^b f(t)dt = \int_0^b f(t)dt - \int_0^a f(t)dt = F(a) - F(b)$

Ok, here's what I meant:

I was looking at $\int_a^b f(t)dt = \int_0^b f(t)dt - \int_0^a f(t)dt = F(a) - F(b)$

and I've been taught that F(x), or F(a) or so on, in this context, is an indefinite integral. Like this:

$$\int f(x) dx$$

But that didn't make sense to me because I don't see how you could evaluate that up to a (in the case of F(a)). To evaluate up to a value you'd need to evaluate from some lower limit to an upper limit, which don't appear on the indefinite integral.

So I was trying to ask what F(x) is. Your definition of it as

F(x) = \int_0^x f(t) dt

makes much more sense.

I'm a little confused about the difference between an antiderivative and an indefinite integral. Can you please explain?

Thanks.

 

[HallsofIvy]

Ok, here's what I meant:

I was looking at $\int_a^b f(t)dt = \int_0^b f(t)dt - \int_0^a f(t)dt = F(a) - F(b)$

Now, this is wrong! The final term should be F(b)- F(a). Perhaps a typo?

and I've been taught that F(x), or F(a) or so on, in this context, is an indefinite integral. Like this:

You may remember it that way but I doubt you were taught that. What is true is that F(x)= \int_0^x f(t)dt is an indefinite integral but F(a)= \int_0^a f(t)dt is a definite integral.

$$\int f(x) dx$$

But that didn't make sense to me because I don't see how you could evaluate that up to a (in the case of F(a)). To evaluate up to a value you'd need to evaluate from some lower limit to an upper limit, which don't appear on the indefinite integral.

The upper limit does appear on the indefinite integral- it is the variable x. To evaluate at, say, x= a, gives a number- the definite integral.

So I was trying to ask what F(x) is. Your definition of it as

F(x) = \int_0^x f(t) dt

makes much more sense.

I'm a little confused about the difference between an antiderivative and an indefinite integral. Can you please explain?

They are "almost" the same thing. Though it would be better to say "an antiderivative" and "the indefinite integral". If, for example, f(x)= 2x, then the "indefinite integral" is \int 2xdx= x^2+ C. Notice the "+C"? Choosing a specific "C" give one of the many possible antiderivatives, functions whose derivative is 2x.

Thanks.

 

[Flumpster]

Now, this is wrong! The final term should be F(b)- F(a). Perhaps a typo?

Absolutely a typo. :) sorry!

You may remember it that way but I doubt you were taught that. What is true is that F(x)= \int_0^x f(t)dt is an indefinite integral but F(a)= \int_0^a f(t)dt is a definite integral. The upper limit does appear on the indefinite integral- it is the variable x. To evaluate at, say, x= a, gives a number- the definite integral.

You'd be surprised at the quality of the instruction here - for example, one of my teachers didn't know what this

f(x+h)-f(x)
----------
h

lim h--> 0was.

I understand that the upper limit is x, I just mean that that I would expect the definition of the indefinite integral to contain *some* reference to a lower bound of 0, which does in fact appear in the definition of the indefinite integral you guys gave (and does *not* appear in any definition I've been given). Otherwise the notation doesn't make sense. It would be like trying to evaluate

\int_{}^x f(x) dx

They are "almost" the same thing. Though it would be better to say "an antiderivative" and "the indefinite integral". If, for example, f(x)= 2x, then the "indefinite integral" is \int 2xdx= x^2+ C[/tex]. Notice the "+C"? Choosing a specific "C" give one of the many possible antiderivatives, functions whose derivative is 2x.

<br /> <br /> That makes sense. It&#039;s like the indefinite integral describes a group of functions that are antiderivatives. Is that correct?<br /> <br /> Also, can you say,<br /> <br /> \int f(x) dx = F(x) + C<br /> <br /> where F(x) is an antiderivative?<br /> <br /> Thanks :) :)

 

[SammyS]

...
That makes sense. It's like the indefinite integral describes a group of functions that are antiderivatives. Is that correct?

The indefinite integral describes a family of functions each of which is an antiderivatives of the integrand, & these function differ from each other by at most a constant term (i.e. an additive constant).

Also, can you say,

\int f(x) dx = F(x) + C

where F(x) is an antiderivative?

Thanks :) :)

You say,
\int f(x) dx = F(x) + C
where F(x) is an antiderivative of f(x) .

 

[Flumpster]

Thanks a lot everyone!

:D