Asked by Darren
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
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
Answers
Answered by
Steve
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)]
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)]
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.