##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##

  • Thread starter happyparticle
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In summary: If you are given ##F_x(x,0)## and ##F_y(1,y)##, then the first term on the left is the horizontal component of \mathbf{F}_x and the second is the vertical component.
  • #1
happyparticle
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Homework Statement
Integral
Relevant Equations
##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##
Hi,
This is the first time I see this kind of integral. I'm not sure how to resolve it.

##
\int_0^1 F \cdot dr
##

##
\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy
##

##F = (y,2x)##
I don't know the values of ## F_x(x,0) ## and ## F_y(1,y)##
 
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  • #2
EpselonZero said:
Homework Statement:: Integral
Relevant Equations:: ##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##

Hi,
This is the first time I see this kind of integral. I'm not sure how to resolve it.

##
\int_0^1 F \cdot dr
##

##
\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy
##

##F = (y,2x)##
I don't know the values of ## F_x(x,0) ## and ## F_y(1,y)##
It is unclear what of the above you are given and what is your own work. Where did ##F = (y,2x)## come from?
You should be able to perform the integral ##\int_0^1 F_x(x,0)dx##. (I assume you realize it means ##\int_0^1 \frac{\partial F(x,0)}{\partial x}dx##.)
 
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  • #3
I don't have anymore information. F is a force.
This is a scalar product, but I don't know the value of ##F_x##
 
  • #4
haruspex said:
It is unclear what of the above you are given and what is your own work. Where did ##F = (y,2x)## come from?
You should be able to perform the integral ##\int_0^1 F_x(x,0)dx##. (I assume you realize it means ##\int_0^1 \frac{\partial F(x,0)}{\partial x}dx##.)
Unless [itex]F_x[/itex] indicates the horizontal component of [itex]\mathbf{F}[/itex], which is given as [itex]y[/itex].
 
  • #5
That's what I thought, but the answer should be 0 + 2, but I don't get that.

##(y,0)\cdot(x,0) \neq 0##

##(0,2x)\cdot(1,y) \neq 2##
 
  • #6
EpselonZero said:
That's what I thought, but the answer should be 0 + 2, but I don't get that.

(y,0)⋅(x,0)!=0

(0,2x)⋅(1,y)!=2

The question is asking you to sett [itex]x = 0[/itex] in the integral with respect to [itex]x[/itex]: [tex]
\int_0^1 F_x(x,0)\,dx = \int_0^1 0\,dx.[/tex] Similarly, the question is asking you to set [itex]x = 1[/itex] in the integral with respect to [itex]y[/itex].
 
  • #7
I'm still confuse. Why not y = 0 in the first integral?

In the second integral, I understand x = 1 so 2x = 2 * 1, but what about x? I'm not sure to understand what respect to means.

For me, ##F_y(1,y)## means ## x = 1## and ##y = y## so ## y = y ## and ## x = 2##

At this point, I'm so confuse I'm not even sure what I'm typing.

Basically, the first integral is on the x-axis and the second on the y axis. ##\int_0^1 F \cdot dr + \int_0^1 F \cdot dr## = ## \int_0^1 Fx \cdot (x,0) dx + \int_0^1 Fy \cdot (1,y) dy##

I thought ##Fx## = 1 so ##1(x,0) = x## and ##2x(1,y) = 2x + 2xy##
 
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  • #8
EpselonZero said:
Why not y = 0 in the first integral?
I think that's what @pasmith meant to write.
 

Related to ##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##

1. What is the purpose of the integral "##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##"?

The purpose of this integral is to calculate the total change in the function F along the x-axis from 0 to 1, and the total change in the function F along the y-axis from 0 to 1. This can help in understanding the overall behavior of the function in the given range.

2. How is the value of the integral "##\int_0^1 F_x(x,0)dx + \int_0^1 F_y(1,y)dy##" calculated?

The value of this integral is calculated by dividing the given range (0 to 1) into smaller intervals and approximating the function F at each interval. The sum of these approximations is then multiplied by the width of each interval to get the total change in the function along the x-axis and y-axis.

3. Is this integral a definite or indefinite integral?

This is a definite integral as it has a specific range of integration (from 0 to 1) and will result in a single numerical value.

4. What are the applications of this integral in science?

This integral has various applications in science, particularly in physics and engineering. It can be used to calculate work done, displacement, and other physical quantities in systems with changing forces or fields. It can also be used to analyze the behavior of a function in a given range, which can be useful in predicting patterns or trends.

5. Are there any limitations to this integral?

Yes, there are limitations to this integral. It can only be used for continuous functions and may not accurately represent the behavior of discontinuous or highly oscillatory functions. Additionally, the accuracy of the integral depends on the number of intervals used for approximation, so a larger number of intervals may be required for more complex functions.

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