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I have trouble understanding easy integral concept.

Here, one of the integral theorem states:

Integral Theorem

And I was wondering what would happen if $F'(x)$ is back in to the integral.

For example, $F(x) = \int_{-1}^{x}\ {2t+t^2}, x \in [-1,5]$

$$F'(x) = 2x + x^2, x \in (-1,5)$$

Then if $F'(x)$ is back in to $\int_{-1}^{x}$ as in $\int_{-1}^{x}\ {2x+x^2}$, should it return $F(x)$?

I mean is the following statement true?

$$\int_{-1}^{x}\ {2x+x^2} = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}$$

What I am confused about is the change of the variable above.

Jeel Shah
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Belphegor
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2 Answers2

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I mean is the following statement true? $$\int_{-1}^{x}\ {2x+x^2}\,dx = \int_{-1}^{x}\ F'(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$$

Not quite: $\text{If}\; F(x) = \int_{-1}^x \,2t + t^2 \,dt$, while it is true that $F'(x) = 2x + x^2$, we cannot write, meaningfully, $$F(x) = \int_{-1}^\color{blue}{\bf x} 2\color{blue}{\bf x} + \color{blue}{\bf x}^2\,d\color{blue}{\bf x}$$

The reason $t$ is used in the original integral $F(x) = F(x) = \int_{-1}^{x}\ {2t+t^2}\,dt$ is merely to stand in as a "dummy variable" which is needed only until we compute the indefinite integral, evaluate at the bounds, which thus gives us a function of $x$: $F(x)$. $F'(x)$ is also a function of $x$.

What you can say is that $F(x) + C = \int F'(x) \,dx$, since here we are not using bounds of integration that are nonsensical.

We can integrate the derivative of a function, and we can find the derivative of the integral of a function, yes indeed. One can think of them, informally, as inverse processes given a function like the one you've posted.

amWhy
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There's at least one problem with this: you have an expression of the form $$\int_c^x F'(x) \, dx,$$ and the variable occurs as the variable of integration and in the limits, which does not make sense. So if you change the name of one of the variables here, the question makes more sense.

But the answer is yes: $$\int_c^\xi \left( \frac{d}{dx} \int_c^xf(t)\,dt \right) \, dx= \int_c^\xi f(x) \, dx.$$

Clarification: The issue with the variables doesn't have to do with properties of the derivative at all. I am told that in the subject of mathematical logic, the issue as described as "free variable vs. bound variable". If you are evaluating $$\psi(\xi) = \int_c^\xi \varphi (x) dx,$$ it does not make sense call $\xi$ by the same name as $x$, because for each $\xi$ that you plug in, you need $x$ to vary over the interval of integration. It doesn't make sense to "plug in" a particular $x$ into the function $\psi$.

Eric Auld
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  • Actually it makes sense: $∫_c^x f(x)dx = ∫_c^x (x \mapsto f(x)) = ∫_c^x f$. Inner $x$ can be thought as “local variable”. – user87690 Aug 10 '13 at 16:32
  • @Eric Auld so you mean once you get F'(x) from F(x), you cannot put F'(x) back to the integral where it is derived from? – Belphegor Aug 10 '13 at 16:33
  • @user87690 I don't see the sense in this. If you plug in say $x=3$, then you are integrating a constant function with respect to a constant? That doesn't make sense. – Eric Auld Aug 10 '13 at 16:34
  • @Belphegor I added some to my answer; see if that clarifies a bit. – Eric Auld Aug 10 '13 at 16:40
  • @EricAuld: If I plug $x = 3$, I get $∫_c^3 f(x)dx$ since plugging affects only free occurences of a variable. The inner occurence is bound. It's like plugging $x = 3$ into $(x = 0) ∨ ∀x(x > 0)$ you get $(3 = 0) ∨ ∀x(x > 0)$. I agree that using the same variable is not good choise for readability, but it's not a nonsense. – user87690 Aug 10 '13 at 16:43
  • @user87690 The integral is an operator on functions, so notation wise $\int_c^x f$ makes the most sense. But many functions are in terms of some formula so then $\int_c^x f(t) , dt$ is more useful. But $t$ is used only to write the function as a formula, for example for $f(t) = t^2$. It does not stand for a number but only as a means to specify the function. Whereas $c$ and $x$ are actual numbers. So $\int_c^x f(x) , dx$ wouldn't make any sense. Also, I don't understand what $(x = 0) \lor \forall x(x > 0)$ is supposed to mean. – Pratyush Sarkar Aug 10 '13 at 17:28
  • @PratyushSarkar: I 100% agree with you until “So $∫_c^x f(x)dx$ wouldn't make any sense”. Contrary, since $f(x)dx$ is just traditional notation for $(x \mapsto f(x))$ in context of integral, the name of inner variable is irrelevant since it doesn't exist any more when the actual one-(nameless)variable function is passed to the operator. $(x = 0) ∨ ∀x(x > 0)$ is just an example of logical formula where the same variable has both free and bound occurence, which is an analogy of $∫_c^x f(x)dx$. – user87690 Aug 10 '13 at 19:52
  • @user87690 So are trying to say (using programming vocabulary) that the $x$ in $\int_c^x$ is outside the scope of $f(x) , dx$ and hence it is justified to use it as the upper limit also? In that case the $x$ in $\int_c^x$ is different from the $x$ in $f(x) , dx$. For $(x = 0) \lor \forall x(x > 0)$, from what I know from logic, I would read it as "$x = 0$ or for all $x$, $x > 0$" (where of course the second part of the statement is false). So are you really allowed to use the same variable for the second part? Seems strange to me. What if you need to use the bounded $x$ again? – Pratyush Sarkar Aug 10 '13 at 22:25
  • @PratyushSarkar: Yes. Of course that $x$ in $f(x) dx$ is different from that in $∫_c^x$. If it was the same $x$ then it's wouldn't make sense. $∀x, x>0$ is just a formula. Alone it's neither false or true. It's false in the structure of real numbers with stadard interpretation of $>$. The point is that when you substitute $x = 3$, you get $(3 = 0) ∨ ∀x(x > 0)$ and not $(3 = 0) ∨ ∀3(3 > 0)$. You substitute only into free occurences of variable. I used is as counterargument to “$∫_c^x f(x)dx$ makes no sence since if you plugged $x = 3$ you would get $∫_c^3 f(3)d3$”. – user87690 Aug 11 '13 at 12:21