# I Is this a Definite or Indefinite Integral?

1. Aug 17, 2016

### Valour549

$$F(x) = \int_a^x f(t) dt$$
I have found various arguments online for both.

Personally I think it's an indefinite integral because:

1) Its upper limit is a variable and not a constant, meaning the value of the integral actually varies with x. This is no different to the family of primitives represented by the indefinite integral, which also varies with x. $$\int f(x) dx$$

2) If F is actually a definite integral then its value must be a constant, which in turn means its derivative must be zero, yet the derivative of F(x) is actually f(x) according to the Fundamental Theorem of Calculus.

Also, I found this but I think it brought more confusion than clarity.

2. Aug 17, 2016

### PeroK

It's a definite integral with a variable upper limit.

3. Aug 17, 2016

### Staff: Mentor

I agree with @PeroK that it is a definite integral whose upper limit is a variable.

4. Aug 17, 2016

### dextercioby

I believe the definition of an indefinite integral excludes the presence of integration limits (be they given and constant numerical values, of functions), therefore it follows that the mathematical object in the RHS of the first line of the original post above must be a definite integral.

5. Aug 17, 2016

### chiro

Hey Valour549.

An indefinite integral is an integral without limits. A definite integral is an integral with limits that can be evaluated.

That is the standard definition of both an indefinite and a definite integral.

6. Aug 18, 2016

### pwsnafu

For each x, you are calculating the area under f over the interval $[a,x]$, hence it is clearly a definite integral.
Doing this for each x, and then defining a function from this computation, does not change this fact.

I disagree with the scanned text. F is "antiderivative of f" but not an "indefinite integral". The latter should be reserved for $\int f(t) \, dt$ which is an operator from a function to an equivalence class of function. See this.

7. Aug 18, 2016

### Valour549

What would you call the LHS of the first line then? According to the scanned text (in the spoiler), it calls the LHS function "an indefinite integral of the integrand [on the RHS]". Do you agree with this?

Because succinctly put, that quote basically says "The definite integral $\int_a^x f(t) dt$ with a varying upper limit defines a function which is an indefinite integral", which is kind of ridiculous.
I was going to argue that since you're calculating the area under $f$ over the interval $[a,x]$, regardless of which $x$ chosen (as long as it's a real number), the result is going to be a numerical value, which when differentiated always gives zero. Yet FTC states the derivative of $\int_a^x f(t) dt$ is $f(x)$, which is certainly not zero.

But then I read this. So it would appear that despite being a definite integral, as you claim, the derivative (w.r.t to $x$) of $\int_a^x f(t) dt$ is actually not zero by virtue of the fact that $x$ exists in its upper limit. But to be fully honest, I still can't get my head around what is wrong with my argument above!

The link you provided seems beyond my scope. But I noted that the 8th line of the scanned text says "ONE of the antiderivatives is...", which is why I said it brought more confusion than clarity.

8. Aug 18, 2016

### PeroK

As was pointed out in post #6, an indefinite integral is an equivalence class of functions (differing by a constant). What you have is a straight function of $x$ generated by evaluating a definite integral with a variable upper limit.

9. Aug 18, 2016

### pwsnafu

You are not understanding the difference between integrating over $[a,x]$ for fixed constants a and x, versus integrating over $[a,x]$ while keeping a fixed and letting x vary. For a fixed x (e.g. $x=16$), you have a definite integral which is a real number, and your intuition is right. But that is not what you are doing.
For each $x$ you are defining $\int_a^x f(t)dt$. You are then defining the function $x \mapsto \int_a^x f(t) dt$. That is not a constant.

You seem to believe that "definite integral is a constant". Nowhere in the definition of definite integral does it say that. The definite integral depends on the function integrated and the terminals. If you vary the function (such as sequences of functions) you vary the integral. If you vary the terminals (like what you are doing here) you vary the integral.

Edit: edited because what I wrote could be more harsh than I first thought.

Last edited: Aug 18, 2016
10. Aug 19, 2016

### Valour549

Yeah, in this respect I'm now convinced that $\int_a^x f(t) dt$ is a definite integral.

The definite integral $\int_a^x f(t) dt$ = $F(x)$, a single function- a particularly useful antiderivative of $f(x)$.

The indefinite integral$\int f(x) dx$ = $F(x)+C$, a class of functions- the family of all antiderivatives of $f(x)$.

Is that correct?
In this respect, I'm also convinced that that $\int_a^x f(t) dt$ is a definite integral.

Because I remembered that the derivative of a point actually has nothing to do with the numerical value at that point.

You see I thought about the basic function $f(x) = x$. If we pick any real number for $x$, then $f$ takes on that particular number. Of course the number changes depending on the $x$ we choose, but it's still always going to be a "number". And the derivative of any "number" is zero, so $\frac{d}{dx}f(x) = 0$. See how this flawed argument is, in essence, identical to the one I was having trouble with?

Of course the derivative of $f(x) = x$ is 1, as it measures the rate of change, and really has nothing to do with the value of $f(x) = x$ at that point. In the same way, $\frac{d}{dx}\int_a^x f(t)dt$ measures the rate of change of this "area-thus-far" function, and the fact that $\int_a^x f(t)dt$ takes on different numerical values for different choices of $x$ tells us that its rate of change is certainly not zero.

11. Aug 19, 2016

### PeroK

Yes, exactly.