Okay, here's a tricky one about the Applications of the Integral. Find a constant c between 1 and 9 such that the average value of the function f(x) on the interval [1,9] is equal to f(c)....where do you start??

Question

anonymous · Answer

Remember that the average value of a function f(x) is $$1 \div (b-a) * \int\limits_{a}^{b} f(x)dx$$. Is that right?

anonymous · Answer

yes

anonymous · Answer

We're going to be setting something equal to something else, I think, but I don't know how to set it up...

hero · Answer

I wouldn't express the average value from a to b in that form, but ...
Hint: c is between a and b

anonymous · Answer

Oh! So split the integral!

hero · Answer

:D

anonymous · Answer

One moment.

anonymous · Answer

Oh, I forgot: $$f(x)=6x^2-4x+2$$ What about the average part? And the f(c)? This is what I'm thinking: $$\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx$$

anonymous · Answer

Set it equal to f(c)?

anonymous · Answer

a=1, b=9?

anonymous · Answer

So, $$\int\limits_{1}^{c}[6x^2-4x+2] dx+\int\limits_{c}^{9}[6x^2-4x+2] dx=6c^2-4c+2$$

anonymous · Answer

Hey...Still didn't get the right answer. Is this correct?