# Second Fundamental Theorem of Calculus Question

• ASmc2
In summary, we use the first fundamental theorem of calculus to differentiate the integral of a function f(t) with respect to x, where x is the upper limit of integration. This is because when we integrate f(t) from a to x, we end up with a function of x, and the only variable we can differentiate with respect to is x. However, if we have f(x) under the integral, we can still use the same notation but it is important to distinguish between the function and the value of the function, and to only use the variable of integration inside the integrand.
ASmc2
Why do we say that f is a function of t and then take the derivative of the integral with respect to x? This confuses me because x is also the upper limit of integration.

Can you please state your version of the second fundamental theorem? My version (the one found on wikipedia) doesn't have your variables in it. It would be easier to explain if I know what we have.

Ok.
I have the expression the integral of f(t) with respect to t from a to x.
The derivative of this with respect to x is f(x).
This is true for every x chosen.

I assume you mean something along the lines of
$$\frac{d}{dx}F(x) = \frac{d}{dx}\int_a^x f(t)dt$$

Which is the http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part".

When integrating $f(t)$ from $a$ to $x$, you end up plugging $x$ and $a$ in for $t$.
$$\int_a^x f(t)dt = F(x) - F(a)$$
$$\frac{d}{dx} \int_a^x f(t)dt = \frac{d}{dx} [F(x) - F(a)]$$
Since $F(a)$ is a constant, you drop it when you differentiate.
$$\frac{d}{dx} [F(x) - F(a)] = \frac{d}{dx} F(x) - \frac{d}{dx} F(a) = \frac{d}{dx} F(x) - 0 = \frac{d}{dx} F(x) = f(x)$$When you differentiate, you have to do so with respect to $x$, because $t$ has been integrated out of the formula. Basically, this is just changing from one variable to another, the only difference being wherever you saw $t$, you replace with $x$.

A specific example of this theorem in action is with $ln\;x$:

$$ln\;x = \int_1^x \frac{dt}{t}$$

Does this help?

Last edited by a moderator:
ASmc2 said:
Why do we say that f is a function of t and then take the derivative of the integral with respect to x? This confuses me because x is also the upper limit of integration.
If F is an anti-derivative of f, then
$$\int_a^b f(t)dt= F(b)- F(a)$$
the "definite integral".

Similarly,
$$\int_a^x f(t)dt= F(x)- F(a)$$
so we now have a function of x. The only thing we can differentiate with respect to is x.

Thank you. After I have posted this thread, I proved to myself that the statement makes sense by using the First Fundamental Theorem (which is what you guys are saying). But now I have another question:
What if we have f(x) instead of f(t) under the integral and we take the integral with respect to x? Would that make the statement bogus?

ASmc2 said:
Thank you. After I have posted this thread, I proved to myself that the statement makes sense by using the First Fundamental Theorem (which is what you guys are saying). But now I have another question:
What if we have f(x) instead of f(t) under the integral and we take the integral with respect to x? Would that make the statement bogus?

I think it would be (I'm not sure, though).
$$\int_a^x f(t)dt$$
When you plug into the antiderivative, $F(t)$, you're plugging in the values of $t$ that you're integrating over, from $a$ to $x$. The way I interpret it (which may be mistaken since I'm only an undergraduate studying math) is that you're plugging in functions $t(x) = a$ to $t(x) = x$, and so when you integrate, you plug in those functions of $x$.
However, when you're integrating
$$\int_a^x f(x)dx$$
you're essentially calling $x$ a function of itself, since you'd be plugging in $x(x) = x$.

Maybe someone that's more knowledgeable on this than myself will comment.

You could use the same sign for the variables, but they are not the same. It isn't "bogus", you just need to know which belong to which operator. Of course, it is more instructive to use different signs for different variables.

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It is important to distinguish between a function and the value of a function. Given f is a function then f(x) is a value of that function, and so too is f(t) or f(42) etc. When we define a function we define the action resulting in its value and the variable we use is arbitrary. Thus for example saying f(x) = 2x+3 defines the function f in exactly the same way as saying f(z) = 2z+3 or f(t) = 2t+3. The use of different variable names is immaterial. The use of y=f(x) is a means of defining the graph of the function as a set of points (x,y) satisfying y=f(x) or equivalently the set of points (t,f(t)) (z,f(z))...
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In the notation for definite integrals (which are values, not functions) the variable of integration is meaningful only inside the the integrand. It is exactly the same as summation notation where the index should only appear inside the summand.
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So take a function f and define another function A from it by defining the value of that function using a definite integral involving f:

$A$ is the function mapping the number $x$ to the value $A(x)$ given by the definite integral:
$$A(x) = \int_a^x f(t)dt$$
Then the fundamental theorem states that the derivative of A is the function f, i.e. A' = f.

Note that it is NOT correct to say that "the derivative of A(x) is the function f(x)" which is trying to take the derivative of something which is NOT a function and identifying it with another something which is NOT a function.

Now we can speak of the derivatives of variable expressions but only when we are thinking of those expressions as functions of a specific independent variable. Thus we can speak of the derivative of f(x) with respect to x, and similarly we can speak of the derivative of f(g(x)) or x f(x) w.r.t. x.
We are using a function (or functions) to write and expression in x which is itself a function of x.

 Thus it is OK to say: "The derivative of A(x) with respect to x, is f(x)". [end edit]

It's a subtle distinction but if ignored it leads to the confusion expressed in the O.P.

Just want to make a quick point: the following expression:

$$\int_a^x f(t)dt$$

is not a function of t, and is a function of x. We can't, for example, evaluate the expression at t=5 (or at any other value of t, for that matter). Whereas we can evaluate it when x=5.

So it is nonsensical to take "the derivative with respect to t". However, it is valid to differentiate it with respect to x.

OK. Now I understand. Thank you all, especially jambaug.

ASmc2 said:
Why do we say that f is a function of t and then take the derivative of the integral with respect to x? This confuses me because x is also the upper limit of integration.

x is a variable so the integral is a function. It is this function that you are differentiating.

## What is the Second Fundamental Theorem of Calculus?

The Second Fundamental Theorem of Calculus is a mathematical theorem that relates the process of differentiation to the process of integration. It states that if f(x) is a continuous function on the interval [a,b] and F(x) is the anti-derivative of f(x), then the definite integral of f(x) from a to b is equal to F(b) - F(a).

## How is the Second Fundamental Theorem of Calculus different from the First Fundamental Theorem of Calculus?

The First Fundamental Theorem of Calculus states that the integration of a function can be calculated by finding its anti-derivative. The Second Fundamental Theorem of Calculus, on the other hand, relates the derivative and integral of a function, showing that they are inverse operations.

## What are the applications of the Second Fundamental Theorem of Calculus?

The Second Fundamental Theorem of Calculus has many practical applications in physics, engineering, and other fields. It can be used to find the displacement, velocity, and acceleration of an object, calculate the area under a curve, and solve various optimization problems.

## Can the Second Fundamental Theorem of Calculus be applied to all functions?

No, the Second Fundamental Theorem of Calculus can only be applied to continuous functions. If a function is discontinuous or has a vertical asymptote within the interval of integration, then the theorem cannot be applied.

## What is the proof of the Second Fundamental Theorem of Calculus?

The Second Fundamental Theorem of Calculus can be proved using the Fundamental Theorem of Calculus, the Mean Value Theorem, and the definition of the derivative. The proof involves showing that the anti-derivative of a function is the same as the derivative of its integral, and then using the Mean Value Theorem to demonstrate that the difference between the two is equal to the original function.

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