# Is the definite integral of the function necessarily the anti-derivative?

Let's say you have a function defined as $$g(x)=\int_1^xf(t)dt$$

By the integral definition, g(x) is the area under the curve of f(x) from 1 to x.

eg: g(5) is the area under f(x) from 1 to 5.

I want to make the distinction that g(x) is NOT defined as an indefinite integral like $$g(x)=\int f(t)dt$$

Now, let's say I am given some arbitrary graph of f.

With the first definition (definite integrals), is g(x) still considered the anti-derivative of f(x)? This matters when exploring ideas about concavity and increase/decrease, and needing to examine the derivatives of g(x). Based on the F.T.C., I know that the derivative of the integral is the function itself. Can we say g'(x) = f(x) and that g''(x) = f'(x)? I know I can say this if g(x) was defined the 2nd way, with indefinite integrals.

But, can I say the same thing when given the limits of integration of $\int_1^x f(t)dt$ ? Or, is g(x) now specifically defined strictly as the area under the curve of f(x) from 1 to x? The 2 different types of integrals (definite vs indefinite) seem to mean different things. Only the 2nd one seems to fit into the antiderivative model (vs. being a specifically defined function.) I guess I am not seeing the connection b/w the two. Let me know if I am unclear.

• The integral is not defined as the area under a curve. $g$ is an antiderivative of $f$ if $f$ is continuous: otherwise FTC does not apply; indeed, it is not hard to construct functions where $g(x)=\int_a^x f(t)\,\mathrm{d}t$ but $g'(x)\neq f(x)$. – Julien Godawatta Mar 27 '14 at 21:53

Not entirely sure I am understanding you correctly, but I think that what this question comes down to is the definition of the integral.

It seems as if you are saying that the integral is defined to be the area under a curve. Now, this is a good approach for a first course in calculus, but for more advanced work, the integral is defined differently and then we do it the other way around: the area is defined to be the integral (in suitable cases - I am simplifying the argument here).

The point is, how are you going to define the area of a figure if you don't use an integral? It's easy to have a "gut feeling" for area as "the amount of space in a region", but if you try to define it more carefully, and to do so in a way that enables one to calculate the areas of specific figures, then I think you will agree that it's not so easy after all. For example, you might say: draw the figure on graph paper and then count the squares. But then you are faced with questions such as: how big should the squares be? What do you do when there are squares partially outside the region? And perhaps most important of all, how do you know that you can always do this?

There are various ways to define what is meant by an integral. Probably the first you will meet is the Riemann integral. If you define it this way it is possible after some work to prove the (first) fundamental theorem of calculus.

Theorem. If $f$ is a continuous function on a closed interval $[a,b]$, and if $$F(x)=\int_a^x f(t)\,dt\ ,$$ then $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $F'(x)=f(x)$ for all $x$ in $(a,b)$.

Hope this helps.

• Ok, so given that integral definition of F(x), if the conditions are met, we can then conclude that F'(x) = f(x) ? In other words, yes, you can treat F(x) as the anti-derivative of f(t), and not just an area function? – JackOfAll Mar 28 '14 at 0:09
• That's right. If you do the theory of integration carefully, then "area" is a secondary consideration and not the primary definition. Another point: as you may know, you can use integration to solve other geometric and physical problem such as volume and mass. If your concept of an integral is purely one of "area", it's a bit hard to see how these other applications come about. – David Mar 28 '14 at 1:02

You're hitting on a really important question here. By "definition", $\int_1^x f(t) dt$ is the area under the graph of $f$ from $1$ to $x$ (I put definition in scare quotes because we haven't really defined what area means, but let's accept that to be an understood idea), so it is a specific function.

On the other hand an antiderivative of $f$ by definition is any function $g$ whose derivative is $f$. So $f$ has a whole class of antiderivatives. At least for continuous functions $f$, the notation $\int f(t) dt$ by definition means the class of antiderivatives of $f$.

Now, suppose that $f$ is continuous (this is critical). Then the FTC tells you that the first specific function is one of the functions in the whole class of functions that make up $\int f(t) dt$.

So, to answer one of your questions, if you are given an arbitrary continuous $f$, you can't just choose to consider the "first" $g$ to be an antiderivative of $f$: this first $g$ already has a definition involving areas. The miracle of the FTC is that it turns out that this first $g$ is an antiderivative for $f$.

This fact isn't totally unintuitive: it's basically just saying that the rate at which the area is changing is equal to the height of the function. You can verify this directly with constant and then linear functions. You can vaguely convince yourself that the logic is the same for arbitrary continuous functions, but without a rigorous definition of area, you can't do any more than that.

• So the bottom line is that if the conditions are met, we can then conclude that g(x) is the anti-derivative of f(t), and not just an area function? – JackOfAll Mar 28 '14 at 0:10
• Yes! As long as you replace "the anti-derivative" with "an anti-derivative". – Abram Lipman Mar 28 '14 at 0:50

Fundamental theorem of calculus: if $f$ is continuous then $$F(x)=\int_a^x f$$ is a primitive of $f$.