Compute the average value of f(x) = x/(x+3) over the interval [-a,a], 0 < a < 3.
a. 1 + (3/2a)ln[(3-a)/(3+a)]
b. 2a + 3ln[(3-a)/(3+a)]
c. 3/(9-a^2)
d. 1/(2a)
e. None of the above
x/(x+3) = 1 - 3/(x+3)
Integral is x - 3ln(x+3)
evaluate at a and -1, divide by 2a to get
(a): 1 + (3/2a)ln[(3-a)/(3+a)]