Integral and dx inside function

In summary, the conversation discusses working on a problem involving an integral and using integration by parts. The problem involves finding the area between a given curve and a line segment between two points on the curve. The conversation also mentions using the Shoelace formula, and discusses different graphs and points on them.
  • #1
gabel
17
0
Im working on a problem whit the follewing integral:

[tex]I = \int|f(x+dx)dx|[/tex]

Im trying to use int by parts : [tex]t = x + dx \Rightarrow dt/dx = 1 + ddx/dx = ?[/tex], but i have no idee on what ddx/dx is? I think dx -> konstant? so ddx/dx = 1 ?
 
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  • #2
gabel said:
Im working on a problem whit the follewing integral:

[tex]I = \int|f(x+dx)dx|[/tex]

Im trying to use int by parts : [tex]t = x + dx \Rightarrow dt/dx = 1 + ddx/dx = ?[/tex], but i have no idee on what ddx/dx is? I think dx -> konstant? so ddx/dx = 1 ?
Is this exactly how the problem is presented? It doesn't look right to me.
 
  • #3
I don't think that integral is well-defined. Where does it come from?
 
  • #4
I have to following expression:

dA = 2 f(x) dx + 0.5 | f(x-dx) | dx + 0.5 |f(x+da)|dx
 
  • #5
gabel said:
I have to following expression:

dA = 2 f(x) dx + 0.5 | f(x-dx) | dx + 0.5 |f(x+da)|dx
I'm guessing you're doing numerical integration using the trapezoid rule where each subinterval is of length 2##\Delta x##.

Why do you have absolute values on two of the terms, but not on the third?
Why do you have f(x - dx) in one place and f(x + da) in another?

If you're using the trapezoid rule to approximate an interval, the idea is to pick a value for n and actually add up some numbers, not do an integration.

When you post a question here you're supposed to use the problem template, the first part of which is a complete problem statement. If you had followed the forum guidelines, we wouldn't have wasted so much time trying to figure out what you're trying to do.
 
  • #6
Mark44 said:
I'm guessing you're doing numerical integration using the trapezoid rule where each subinterval is of length 2##\Delta x##.

Why do you have absolute values on two of the terms, but not on the third?
Why do you have f(x - dx) in one place and f(x + da) in another?

If you're using the trapezoid rule to approximate an interval, the idea is to pick a value for n and actually add up some numbers, not do an integration.

I'm using the shoelace formula (http://en.wikipedia.org/wiki/Shoelace_formula) for a triangle, and then letting

x_1 = x - dx , x_2 = x, x_3 = x + dx
y_1 = f(x-dx), y_2 = f(x), y_3 = f(x+dx).

Aften cleanup I am left whit this
dA = 2 f(x) dx + 0.5 | f(x-dx) | dx + 0.5 |f(x+da)|dxNB, latex is not working for me atm.
 
  • #7
gabel said:
I'm using the shoelace formula (http://en.wikipedia.org/wiki/Shoelace_formula) for a triangle, and then letting

x_1 = x - dx , x_2 = x, x_3 = x + dx
y_1 = f(x-dx), y_2 = f(x), y_3 = f(x+dx).

Aften cleanup I am left whit this
dA = 2 f(x) dx + 0.5 | f(x-dx) | dx + 0.5 |f(x+da)|dxNB, latex is not working for me atm.
It's still not clear to me what you're trying to do? Are you trying to find the area of a specific region? If so, you should have numeric values for x1, x2, and so on, as well as for y1, y2, and so on.

Please be more specific about what you're trying to do. From what I can tell, your formulas should NOT involve dx or da and you should not be trying to integrate things as you seem to be trying to do.

LaTeX works just fine, but when you click Post Reply, you should hit the browser's refresh button to see how it will appear. Or you could type your expression in the Preview LaTeX pane to test it first.
 
  • #8
If i have a f(x) = ax+b, given 3 points on f(x), the area from those points shood be zero. Since they will always be on a line. But what heppens if f(x) = sin(x), cos(x), x^2, etc?

Some more details.

Red :f(x) = x
Blue:f(x) = x^2

A_2 = 0, A_1 = something small

Now i want to move A_1 along the red function to se how A_1 behaves.

Im now thing of dx as the distance from A to B along x as dx, A to C as -dx etc for red funciton.

Capture.PNG
 
  • #9
You shouldn't use one letter, f, to mean two different things. The blue graph is, say, f(x) = x. The red graph is, say, g(x) = x2. You have the descriptions reversed.

On a straight line, if you connect any two points with a line segment, any point on the segment will also lie on the line. If you have a curve that isn't linear, and you draw a segment between two points on the curve, some points on the segment will not lie on the curve.

gabel said:
Now i want to move A_1 along the red function to se how A_1 behaves.
What does this mean?
gabel said:
Im now thing of dx as the distance from A to B along x as dx, A to C as -dx etc for red funciton.
A and C aren't on the red graph.The points you show are E, G, and F.

Why are you using the Shoelace formula? That formula gives the area of a polygon. Are you trying to find the area between a given curve and a line segment between two points on the curve?
 
  • #10
Mod note: edited to show what I said and the responses.
Mark44 said:
You shouldn't use one letter, f, to mean two different things. The blue graph is, say, f(x) = x. The red graph is, say, g(x) = x2. You have the descriptions reversed.

On a straight line, if you connect any two points with a line segment, any point on the segment will also lie on the line. If you have a curve that isn't linear, and you draw a segment between two points on the curve, some points on the segment will not lie on the curve.What does this mean?
gabel said:
- I am thinking sort of like convolution.
gabel said:
A and C aren't on the red graph.The points you show are E, G, and F.
gabel said:
- A,B and C are on the blue graph, and the area from that the points A,B,C give is zero. I then want to check how a area constructed on diffrect functions behave. As an example i f=x^2. How will that the area behave?
gabel said:
Why are you using the Shoelace formula? That formula gives the area of a polygon. Are you trying to find the area between a given curve and a line segment between two points on the curve?

- Yes, expect i want to use 3 points.
 
Last edited by a moderator:
  • #11
I don't see how convolution has anything to do with what you're trying to do.

Instead of seeing how an area "behaves" (meaningless because an area doesn't do anything), why not calculate the area between the curve and a straight line that connects two points on the curve. If you have a curve y = f(x) for a <= x <= b, the points (a, f(a)) and (b, f(b)) are on the curve and also determine a line segment.

The equation of the line is y = f(a) + [f(b) - f(a)]/(b - a) * (x - a).

The area between the line and the curve is
$$ \int_a^b |f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a)|dx$$

If the line segment crosses the curve, it's possible for this forumula to result in zero. To avoid that happening, make sure that each line segment lies above the curve or below the curve.
 

1. What is the purpose of having "dx" inside the integral?

The "dx" inside the integral represents the variable of integration. It indicates that the integration is being performed with respect to that variable and is necessary in order to properly calculate the integral.

2. How do you solve integrals with "dx" inside the function?

To solve integrals with "dx" inside the function, you can use techniques such as substitution, integration by parts, or partial fractions. It is important to follow the rules of integration and properly handle the "dx" term in each step of the integration process.

3. What is the significance of the "dx" term in the Fundamental Theorem of Calculus?

In the Fundamental Theorem of Calculus, the "dx" term represents the infinitesimal change in the variable of integration. It is crucial in understanding the connection between differentiation and integration, as it allows us to calculate the derivative of an integral.

4. Can the "dx" term be replaced with a different variable?

Yes, the "dx" term can be replaced with any other variable, as long as it is consistent throughout the integral and the limits of integration. Some common alternatives include "dt", "du", or "dy".

5. Is there a specific order in which to solve integrals with "dx" inside the function?

No, there is no specific order in which to solve integrals with "dx" inside the function. The method used to solve the integral will depend on the specific function and may require the use of multiple techniques. It is important to carefully analyze the integral and choose the most appropriate method for solving it.

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