Fundamental theorem of calculus (something isn't right)

[njama]

Thanks for the explanation jgens. I partially understand what are you saying to me, but I can't really see the big difference. Suppose [-1,x] is interval on which I need to find the area on. So A(x) would be the area. But it can be seen that $\int_{f(x)dx}$=A(x) or A'(x)=f(x). I also checked wikipedia.

Here is what I found:

In calculus, an antiderivative, primitive or indefinite integral[1] of a function f is a function F whose derivative is equal to f, i.e., F ′ = f. The process of solving for antiderivatives is antidifferentiation (or indefinite integration)

 

[jgens]

If you want a simple illustration of how they're different, consider the function $f$ defind by $f(x) = 0$ if $x = 1$ and $f(x) = 1$ otherwise. Now suppose that I want to evaluate $\int_0^xf$. It's a trivial exercise to show that $\int_0^xf = x$. Now if integration were the inverse process of differentiation we would have that $(x)' = 1 = f(x)$ which clearly isn't true. Does this make sense?

 

[njama]

If x=1:
$$\int_0^xf(t)dt= \int_0^x0dt=0(\int_0^x1dt)+C=C$$
If $x\neq 1$
$$\int_0^x1dt=(t+C_1)|_0^x = x-0=x$$
It still valid:
$$F(x)=\left\{\begin{matrix} C, x=1\\ x, x \neq 1 \end{matrix}\right.$$

$$f(x)=F'(x)=\left\{\begin{matrix} 0, x=1\\ 1, x \neq 1 \end{matrix}\right.$$

Edit: This function does not satisfy the theorem, its not continuous, so it fails to work anyway.

 

[elibj123]

If x=1:
$$\int_0^xf(t)dt= \int_0^x0dt=0(\int_0^x1dt)+C=C$$
If $x\neq 1$
$$\int_0^x1dt=(t+C_1)|_0^x = x-0=x$$
It still valid:
$$F(x)=\left\{\begin{matrix} C, x=1\\ x, x \neq 1 \end{matrix}\right.$$

$$f(x)=F'(x)=\left\{\begin{matrix} 0, x=1\\ 1, x \neq 1 \end{matrix}\right.$$

Edit: This function does not satisfy the theorem, its not continuous, so it fails to work anyway.

You cannot just swap f(x) for zero, because the integral "doesn't ask" the function what's its value on a certain point, but what's its value on an interval. Therefore the integral goes through "many" points in which f(x)=1, bumps into a zero, and then goes on. (This is a point where I recommend to read again about integrals, and exercise them, because what you did in the first lines "if x=1 blahblah" show you don't completley understand how to work with them and what do they mean)
But we know that changing a function in a finite number of points, doesn't change its integral, so as far as the integral "knows" he's actually integrating only f(x)=1.

Anti-differentiation, if studied as an operator, must consider any information in f(x), while integration loses this little things that make up f(x).

 

[jgens]

If x=1:
$$\int_0^xf(t)dt= \int_0^x0dt=0(\int_0^x1dt)+C=C$$
If $x\neq 1$
$$\int_0^x1dt=(t+C_1)|_0^x = x-0=x$$
It still valid:
$$F(x)=\left\{\begin{matrix} C, x=1\\ x, x \neq 1 \end{matrix}\right.$$

$$f(x)=F'(x)=\left\{\begin{matrix} 0, x=1\\ 1, x \neq 1 \end{matrix}\right.$$

Edit: This function does not satisfy the theorem, its not continuous, so it fails to work anyway.

No, no, no! This isn't right in the slightest. You can't just integrate using the FTC if $f$ isn't continuous. You need to go back using the definition of the integral like the one that I provided and not what you mistakenly keep taking as the definition of the integral.

Anyway, once you actually do the exercise correctly, you will find that $\int_0^xf = x$; hence, even though $f$ isn't continuous, this integral is. Now, if integration and anti-differentiation were the same thing, differentiating $\int_0^xf$ would show that $f(x) = 1$ which is clearly false. Therefore, anti-differentiation and integration are not the exact same thing.

 

[njama]

OK. You defined the function like this:

And you told me to find:

$\int_0^bf$

Now. If the function is continuous on interval [0,b] then it is integrable on [0,b].

But if function is bounded and have finitely many discontinuity on [0,b] it is still integrable.

So, we can find
$\int_0^bf$

Lets consider any partition on [0,b]. Then either $x_{k}^{*}=0$ or it does not.
If not then
$$\sum_{k=1}^{n}f(x_{k}^{*})\Delta x_{k} = \sum_{k=1}^{n}\Delta x_{k} = b$$
else
$$\sum_{k=1}^{n}f(x_{k}^{*})\Delta x_{k} = -\Delta x_{k} + \sum_{k=1}^{n}\Delta x_{k} = b - \Delta x_{k}$$

which means that the difference between the Riemann sum and b is at most $\Delta x_{k}$. BUT, since $\Delta x_{k}$ approaches zero as $max \Delta x_{k} \rightarrow 0$ it follows that:

$$\int_{0}^{b}f(x)dx = b$$

But the Fundamental Theorem of Calculus Part 2 states that "If f is continious on an interval [0,b] then f has an antiderivative on [0,b]. In particular if a is any number in [0,b] then the function F defined by:
$$F(x)=\int_{a}^{x}f(t)dt$$
is an antiderivative of f on [0,b]; that is F'(x)=f(x) for each x in [0,b], or in an alternative notation:
$$\frac{d}{dx}\left [ \int_{a}^{x}f(t)dt \right ] = f(x)$$

So f must be continious so that we can conclude that f has an antiderivative. If f is not continious anything can follow.

 

[jgens]

So f must be continious so that we can conclude that f has an antiderivative.

Except for the fact that $f$ clearly isn't continuous so the FTC isn't applicable here. The point of this exercise was to illustrate that if $F$ is a function defined by $F = \int_0^xf$, then $F'$ is not necessarily equal to $f$ as you asserted in post number 30 of this thread.