- #1

Cyrus

- 3,150

- 16

Ok guys, about two weeks ago I posted asking about the fundamental theroem of calculus and the use of the dummy variable. Ive had some time to clear my head. I want to take a fresh new approach to it. I have reread what you guys put and I will just write down what I think. Please be patcient and thanks in advance for putting up with my B.S.

Ok, so first things first. Lets start with a standard integral evaluated from a to b.

[tex] \int^b_a f(x)dx [/tex]

In order to do this, we plug in values between a and b that vary, and sum it over the interval. When we do this, we start with the lower bound, a, and plug it in and evalute it. We multiply this buy dx, which is a small piece of the parameter domain, and repeat the process for the next increment. Notice that in this process, dx was the value of a small increment in the parameter domain. It is usually the limit of delta x, as x goes to zero. No where in this definition of dx, did the value of x go anywhere in the x part of dx. For the first value of a, we did not put f(a)da. Because that is now how we defined the use of the symbol dx. dx has a very specific meaning. Although there is an x in the dx, it does NOT mean put in a value where you see the x in the term dx.

Now lets move on to the FTC: Here we have:

[tex] F(x)= \int^x_a f(x)dx [/tex]

Hopefully at this point, we all agree that this notation is wrong. Lets see why. Lets use crosson and mattgrime's point of evaluating this function at the point x=3. Here we have a

[tex] F(3)=\int^3_a f(3)d3 [/tex]

Notice that we had to put in a 3, for the term d3. But now we have an integral where the function ONLY ever takes on one value, f(3). ALSO, we have a term called d3. What is d3? d3 is nothing, its pure garbage. In fact, the way we defined an integral in the first example tells us that the dx part should NEVER EVER have a value in that x part of dx. Even though dx has an x in it, its defined such that you dont PUT a value in that x, PEROID, EVEN IN STANDARD INTEGRALS THAT DO WORK!

Ok, so now what. Since im a smarta$$. Lets say, ok, I know what an integral is and what dx REALLY means, and based on its definition, I have to leave that x in the dx alone. So lets say I CHOOSE to do that.

[tex] F(3) = \int^3_a f(3)dx [/tex]

But now look what I have done. I

The only way to to satisify ALL THREE of these conditions, is to introduce a separate dummy variable. THEN and ONLY then does it give me what I want, an integral that can be though of as an area up to here, where 'here' is the value of x.

(Not to be rude, but this post is really a quesiton for MattGrime, so please let him respond first if you dont mind)

Ok, so first things first. Lets start with a standard integral evaluated from a to b.

[tex] \int^b_a f(x)dx [/tex]

In order to do this, we plug in values between a and b that vary, and sum it over the interval. When we do this, we start with the lower bound, a, and plug it in and evalute it. We multiply this buy dx, which is a small piece of the parameter domain, and repeat the process for the next increment. Notice that in this process, dx was the value of a small increment in the parameter domain. It is usually the limit of delta x, as x goes to zero. No where in this definition of dx, did the value of x go anywhere in the x part of dx. For the first value of a, we did not put f(a)da. Because that is now how we defined the use of the symbol dx. dx has a very specific meaning. Although there is an x in the dx, it does NOT mean put in a value where you see the x in the term dx.

Now lets move on to the FTC: Here we have:

[tex] F(x)= \int^x_a f(x)dx [/tex]

Hopefully at this point, we all agree that this notation is wrong. Lets see why. Lets use crosson and mattgrime's point of evaluating this function at the point x=3. Here we have a

**FUNCTION**. When we normally see a**FUNCTION**, our standard procedure says to put in a value everywhere we see an x. If we choose to evaluate the point x=3, then we would have to do the following:[tex] F(3)=\int^3_a f(3)d3 [/tex]

Notice that we had to put in a 3, for the term d3. But now we have an integral where the function ONLY ever takes on one value, f(3). ALSO, we have a term called d3. What is d3? d3 is nothing, its pure garbage. In fact, the way we defined an integral in the first example tells us that the dx part should NEVER EVER have a value in that x part of dx. Even though dx has an x in it, its defined such that you dont PUT a value in that x, PEROID, EVEN IN STANDARD INTEGRALS THAT DO WORK!

Ok, so now what. Since im a smarta$$. Lets say, ok, I know what an integral is and what dx REALLY means, and based on its definition, I have to leave that x in the dx alone. So lets say I CHOOSE to do that.

[tex] F(3) = \int^3_a f(3)dx [/tex]

But now look what I have done. I

**STILL**get an integral, which is now something I CAN**POSSIBLY**evalute, BUT what does the integral represent? Now it just means im summing f(3), (3-a) times. AGAIN, this is NOT what I had set out to do in the fist place. So EVEN IF I try to be smart about saying, wait a minute, d3 violates the definition of an integral, leave that part alone as dx, I STILL get something that makes no sense.The only way to to satisify ALL THREE of these conditions, is to introduce a separate dummy variable. THEN and ONLY then does it give me what I want, an integral that can be though of as an area up to here, where 'here' is the value of x.

(Not to be rude, but this post is really a quesiton for MattGrime, so please let him respond first if you dont mind)

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