## anonymous one year ago Find the average value of f(x) = sin(x) over the interval [0, π].

1. jim_thompson5910

Average value (A) of a function f(x) over the interval from x = a to x = b $\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx$

2. anonymous

pi/2?

3. jim_thompson5910

In your case, you have a = 0 b = pi f(x) = sin(x) $\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx$ $\Large A = \frac{1}{\pi-0} \int_{0}^{\pi} \sin(x) dx$ $\Large A = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) dx$ $\Large A = ???$

4. jim_thompson5910

what's the integral of sin(x) ?

5. anonymous

-cos

6. jim_thompson5910

yes -cos(x) + C we can drop the +C because we have a definite integral

7. jim_thompson5910

evaluate -cos(x) when x = pi evaluate -cos(x) when x = 0 subtract the results to get ???

8. anonymous

2

9. anonymous

I got pi over 2 but someone told me just 2

10. anonymous

or sorry 2 over pi

11. jim_thompson5910

yeah the integral portion should evaluate to 2

12. jim_thompson5910

which is why A = 2/pi

13. anonymous

okay thank you very much for clearing that up

14. jim_thompson5910

np