# Average value

1. Jun 11, 2008

### chukie

For what value of c>0 is the average value of the graph on [-5,c] equal to zero and why?

Graph is attached.

I thought that that c should be 4 since at that point the areas cancel out. Not sure though. Could someone help me? Thanks!

#### Attached Files:

• ###### graph.bmp
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2. Jun 11, 2008

### konthelion

I don't see a graph since it's still pending approval, but use the Mean Value Theorem(for integrals) to solve for c.

3. Jun 11, 2008

### chukie

4. Jun 11, 2008

### chukie

ive managed to get the graph

5. Jun 11, 2008

### chukie

i also looked at the mean value theorem. it deals with f'(x)=0 ?

6. Jun 11, 2008

### konthelion

Ah, ok. Yes, c=4 since f is a continuous function on [-5,c], then the average value of f from -5 to c = $$\frac{1}{c-(-5)}$\int_{-5}^{c} f(x)\,dx.$$$. Since the average value = 0, then it's just the integral from -5 to 4(since it's a graph, yeah the areas cancel out)

No, check out the MVT for integrals also known as the average value of a function.

Last edited: Jun 11, 2008
7. Jun 11, 2008

### chukie

okay great! thanks so much =)