Question

Find the average value of f(x) = sin(x) over the interval [0, π].

Answer 1

Average value (A) of a function f(x) over the interval from x = a to x = b

\[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\]

Answer 2

pi/2?

Answer 3

In your case, you have

a = 0
b = pi
f(x) = sin(x)

\[\Large A = \frac{1}{b-a} \int_{a}^{b} f(x) dx\]
\[\Large A = \frac{1}{\pi-0} \int_{0}^{\pi} \sin(x) dx\]
\[\Large A = \frac{1}{\pi} \int_{0}^{\pi} \sin(x) dx\]
\[\Large A = ???\]

Answer 4

what's the integral of sin(x) ?

Answer 5

-cos

Answer 6

yes -cos(x) + C

we can drop the +C because we have a definite integral

Answer 7

evaluate -cos(x) when x = pi
evaluate -cos(x) when x = 0

subtract the results to get ???

Answer 8

2

Answer 9

I got pi over 2 but someone told me just 2

Answer 10

or sorry 2 over pi

Answer 11

yeah the integral portion should evaluate to 2

Answer 12

which is why A = 2/pi

Answer 13

okay thank you very much for clearing that up

Answer 14

np