Find c for Average Value of Graph to be 0: Why c=4?

[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!

 

[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.

 

[chukie]

http://img249.imageshack.us/img249/5679/graphws6.th.png [Broken]

 

[chukie]

ive managed to get the graph

 

[chukie]

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

 

[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)

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

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

 

[chukie]

okay great! thanks so much =)