I need help with Calculus and the Average value theorem

Question

anonymous · Answer

$$f(x)=\sin(\pi x)$$ interval  [1,2]

anonymous · Answer

@satellite73

anonymous · Answer

find the average slope by [f(b) - f(a)] / (b-a),  set equal to derivative of sin(pi*x), solve for x

anonymous · Answer

I can't get an answer though :( @billj5

anonymous · Answer

show me what you have so far

anonymous · Answer

ok so to find the avg value you do $$\frac{ 1 }{b-a } * \int\limits_{a}^{b} f(x)dx$$

since a is 1 and b is 2 1/b-a=1 so that doesn't really matter. 
then when I do $$\int\limits_{1}^{2}sin(pi x)dx$$ I get picos(pix) then when I do f(2)-F(1) I get 2pi but I can't solve the equation with that. I don't get a number @billj5

jim_thompson5910 · Answer

the integral of sin(pi*x) is NOT pi*cos(pi*x)

try it again

anonymous · Answer

oh

jim_thompson5910 · Answer

you're thinking of the derivative

anonymous · Answer

ok I got it it's -cos(pix)/pi

jim_thompson5910 · Answer

yep

jim_thompson5910 · Answer

of course +C at the end

but with definite integrals the "+C" cancels out

anonymous · Answer

so f(c) for this case would be -2/pi correct?

jim_thompson5910 · Answer

I'm not sure what you mean by "f(c)"

but the average value on that interval is indeed -2/pi