Need help with integrals!

[Nate]

I'm doing physics homework. Here's the problem and a crappy drawing of the illustration done in paint. Fear my horrendous paint skills. XD;;

The force on a particle, acting along the x axis, varies as shown. Determine the work done by this force to move the particle along the x axis: (a) from x=0.0 m to x=10.0 m; (b) from x=0.0 m to x=15.0 m



Since it's really crappy, I'll go ahead and tell you that the y axis, each mark represents 100 N, and it levels out at 400. For the bottom of the curve, it levels out at -200 N. It's 0 at 10 and 15 m.

So, I know I have to use integrals. But other than knowing that an integral is the opposite of a derivative, I don't know how to do them. I know that the total work is going to be each little bit of work added together, but an integral is a shorter way of doing that, I guess? So how do I do that? I'm confused. Any help would be appreciated.

[SP1]

You will need to develop the equation of the force function (linear) for each segment x = 0-3, 3-7, 7-11, 11-14, and 14-15. If dW = F*dx, then W = INT(F*dx) over the interval. Alternatively, the integral of a function like this is simply the AREA under the curve. That is, instead of using an integral function for x = 0-3, you could just see that it is a triangle of base 3 m and height 400 N or 600 Nm in area. For areas under the X axis, work is negative, so be sure to subtract that area. This is a quick way to check your integral math.

[Technomancer]

Along with what SP1 said, an integral is just a summation with infitesimally small sections. Here, you have some fairly regular sections with well known formulas for their area (triangles and rectangles). So you really just need to sum up the areas under each section (remember, the negative sections will have negative areas). Given a more difficult curve, you can just create arbitrary blocks and sum their approximate area via the trapezoidal rule.

[Mithrandir]

Since I can't tell if the slopes are even, I'll assume they are and give my take on what that would mean.

Above the Line:
If the slope of the line is equal (x/y vs x/-y) for 0-3 and 5-8, then you really have the area of a rectangle defined by 400x8, or 3200.
Below the Line:
If the slope of the line is equal (x/y vs x/-y) for 10-11 and 14-15, then you have another rectange, with area: 200x4, or 1000.

Total work would be across a section of 3200 - 1000, or 2200.

If I'm reading that wrong YMMV.