HallsofIvy said:No, that is not correct. In fact, your entire method is wrong. This is a one dimensional problem (a curve) and you are doing a two dimensional integral. In general, you need to write the integral in terms of a single parameter. Here, since we have y equals a function of x, we can use x itself as that parameter.
On the curve [itex]y= 4- x^2[/itex], [itex]dy= -2x dx[/itex]. So (x- y) dx= (x- 4+ x^2) dx and [itex]xdy= -2x^2 dx[/itex] That is,
[tex]\int_{(2, 0)}^{(1, 3)} (x- y)dx+ xdy= \int_{x=2}^1 (x^2+ x- 4- 2x^2)dx= \int_{x=2}^1 (-x^3+ x- 4)dx[/tex]
A line integral is a type of integral used in calculus to calculate the work done by a force along a curve or path. It involves breaking down the curve into small segments and adding up the work done by the force on each segment.
A regular integral calculates the area under a curve, while a line integral calculates the work done along a curve. This is because a regular integral uses rectangular slices to approximate the area, while a line integral uses line segments to approximate the work.
Finding the amount of work done is crucial in understanding the energy and forces involved in a system. It can help in predicting the motion of objects and determining the efficiency of a process.
The formula for calculating a line integral is ∫F(x,y)·ds, where F(x,y) is the force function and ds is the infinitesimal arc length along the curve. This can also be written as ∫F(x(t),y(t))·|r'(t)|dt, where r(t) is the parametric equation of the curve.
The direction of the line integral is determined by the orientation of the curve. This can be specified by either using a positive direction (from start point to end point) or by specifying the direction of the curve using a vector function.