# Homework Help: Finding the function to integrate

1. Oct 11, 2009

### Carmen12

1. The problem statement, all variables and given/known data

Define the function over the four intervals in order to find the area under the curve from [-4,6].

To do so, find the equations of the line and circle, using the data in the graph.

2. Relevant equations

None?

3. The attempt at a solution

To be honest, there is no attempt at a solution except staring at it for 4 hours. Given a function, I can integrate. But findind the equation based on the graph confuses me. It has been years since I last did such a thing.

2. Oct 11, 2009

### zcd

There are basically 4 shapes defined by semicircles and line segments. Boundaries are at [-4,-1], [-2,0], [2,0], [4,2], [6,0]. The lines should be easy to figure out, and the equation for a circle centered at the origin of radius a is $$\pm\sqrt{a^{2}-x^{2}}$$

3. Oct 11, 2009

### dancergirlie

rather than trying to find a function for the lines, why not just break them up into simple geometrical shapes and find the areas under the curve?
For example:
the area of a circle is pi(r)^2
and the area of a triangle is 1/2(bh)

4. Oct 11, 2009

### Staff: Mentor

There are four functions you need to find from this graph: 3 linear functions and one that represents the lower half of a circle. Going left to right, the first function is defined on the interval [-4, -2]. The function that represents the lower half of the circle is defined on the interval [-2, 2]. The other two functions are defined on the intervals [2, 4] and [4, 6].

If you are given two points on a line, can you find the equation of the line?
If you know the radius of a circle and its center, can you find the equation of the circle?

5. Oct 11, 2009

### Staff: Mentor

The area under this curve is infinite. Doesn't the problem actually ask you to find the area between the curve and the x-axis?

Also, dancergirlie's tip is a good one, and gives the easiest way to find this area, unless you actually have to come up with the functions.