# Average Value Question

1. May 6, 2007

### t_n_p

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

3. The attempt at a solution

Hope you can see what I've written on the top of the page, but basically I want to find the antider. of f(x) from 1 to 4 (from the graph I suppose). So that should be the negative area under the graph plus the little triangle just above the x axis?

2. May 6, 2007

### Gib Z

Break it up into 2 integrals,
$$\int^{3.5}_1 f(x) dx + \int_{3.5}^4 f(x) dx$$

We can see the 1st integral is just the negative of a trapezium, and it is -7/4.
The 2nd integral is just a triangle, area 1/4. So the integral evaluates to -1.5.

A better way to have done it would have been to see the area from 3 to 3.5 and 3.5 to 4 cancel out making the integral $$\int_1^3 f(x) dx$$, which is -3/2 as expected. Turns out you don't see an anti derivative, but if you were really really desperate, you could have written a piece wise definition of f(x) where f(x) = -x+1 for x between 0 and 2, f(x)=-1 for x between 2 and 3, so on, so forth..and done it the long way.

Anyway, we have the integral now, you can do the rest.

3. May 6, 2007

### t_n_p

So basically average value = (1/3)*(-3/2)
= -1/2

4. May 6, 2007

### Gib Z

Exactly. Good Work.

5. May 6, 2007

### t_n_p

*bows down to maths god*