I Question about definite and indefinite integrals

[jonjacson]

First, just to check, I write what I think and let me know if I am wrong:

The definite integral of a function gives us a number whose geometric meaning is the area under the curve between two limiting points.
We can calculate this integral as the limit of the sum of the rectangles and the integral symbol means a sum over all rectangles.

The indefinite integral of the function f(x) is the function primitive F(x) whose derivative is f(x), and here we are using the fundamental theorem of calculus. Why do we use the same symbol?

My question is, How could we calculate the indefinite integral without the fundamental theorem?

Could we try it using the definite integral between the limits 0 and x?

From a geometric point of view, How could we talk about the indefinite integral?

Thanks!

 

[BvU]

Why do we use the same symbol?

Because basically it's the same thing. Compare using the same term $\sin$ for the sine function and for the value of the sine function at some point $x$.

How could we calculate the indefinite integral without the fundamental theorem?

We could create a table of F values from the sum of rectangle areas, but we would still need to state where we start integrating (i.e. the table would list $F(x)-F(x_{\rm start}$ ). So: Yes to your

Could we try it using the definite integral between the limits 0 and x?

Note: this doesn't give you a function in the from of a 'recipe', though.

From a geometric point of view, How could we talk about the indefinite integral?

It tells you how the area under the curve varies with the independent variable

 

[PeroK]

First, just to check, I write what I think and let me know if I am wrong:

The definite integral of a function gives us a number whose geometric meaning is the area under the curve between two limiting points.
We can calculate this integral as the limit of the sum of the rectangles and the integral symbol means a sum over all rectangles.

The indefinite integral of the function f(x) is the function primitive F(x) whose derivative is f(x), and here we are using the fundamental theorem of calculus. Why do we use the same symbol?

My question is, How could we calculate the indefinite integral without the fundamental theorem?

Could we try it using the definite integral between the limits 0 and x?

From a geometric point of view, How could we talk about the indefinite integral?

Thanks!

You seem to be largely answering your own questions here. The only thing I would add is that you could see $F$ as a function which gives you definite integral between two points:

$\int_{a}^{b}f(x)dx = F(b) - F(a)$

And, it turns out that any $F$ where $F' = f$ does the job. And, you call any such $F$ an indefinite integral of $f$. More precisely, the set of functions $\lbrace F: F' = f \rbrace$ is the indefinite integral.

 

[jonjacson]

Because basically it's the same thing. Compare using the same term $\sin$ for the sine function and for the value of the sine function at some point $x$.
We could create a table of F values from the sum of rectangle areas, but we would still need to state where we start integrating (i.e. the table would list $F(x)-F(x_{\rm start}$ ). So: Yes to your
Note: this doesn't give you a function in the from of a 'recipe', though.
It tells you how the area under the curve varies with the independent variable

The indefinite integral gives you the area (if you substitute for particular values), not the rate of change of the area, which is the function itself right?

You seem to be largely answering your own questions here. The only thing I would add is that you could see $F$ as a function which gives you definite integral between two points:

$\int_{a}^{b}f(x)dx = F(b) - F(a)$

And, it turns out that any $F$ where $F' = f$ does the job. And, you call any such $F$ an indefinite integral of $f$. More precisely, the set of functions $\lbrace F: F' = f \rbrace$ is the indefinite integral.

THanks.

 

[BvU]

The indefinite integral gives you the area (if you substitute for particular values), not the rate of change of the area, which is the function itself right?

Yes.
Example: constant function. Indefinite integral: x itself, linear function. Area changes linearly with the independent variable.

 

[jonjacson]

Yes.
Example: constant function. Indefinite integral: x itself, linear function. Area changes linearly with the independent variable.

I think there was a misinterpretation.

When you talked about changed linearly, you talked about the diferent values a function primitive F(x) can take, I thought you meant its rate of change F'(x) , that is why I said the rate of change was the funcion f(x) itself.

I guess!

 

[pwsnafu]

You might be interested in the Risch Algorithm.

 

[Svein]

Just to add to the confusion: In many cases it is possible to evaluate the definite integral but not the indefinite...

 

[jonjacson]

You might be interested in the Risch Algorithm.

Very interesting yes.

Just to add to the confusion: In many cases it is possible to evaluate the definite integral but not the indefinite...

I understand that. Finding an area is different to finding a function whose differential is a given one.

Have I told you anything confusing?

 

[Svein]

Have I told you anything confusing?

Just an expression. Means: "This information that was not asked for and may not be very relevant"