- #1
DavidWi
- 10
- 0
I don't know... I've worked through this problem about ten times, but each time, I get the same answer. It's a kind of easy problem too. I don't konw what my problem with this is.
If f(x) = sin (x/2), then there exists a number c in the interval pi/2 < x < (3*pi / 2 that satisfies the conclusion of the Mean Value Theorem. Which of teh following could be c?
Here's my work...
The mean value theorem is f(c) = 1/(b-a) * int (f(x), x, a, b).
the indefinate integral of sin (x/2) is -2 cos (x/2) + c
c = -2 (cos ((3 * pi / 2) / 2) - cos ((pi / 2) / 2) / pi
c = -2 (cos (3 * pi / 4) - cos (pi / 4)) / pi
c = -2 ( - sqrt (2) / 2 - sqrt (2) / 2) / pi
c = -2 (- 2 sqrt (2) / 2) / pi
c = -2 (- sqrt (2)) / pi
c = 2 * sqrt (2) / pi
but, the answer sheet only gives answers with pi in the numerator (so i know i must be wrong). Can anyone help?
If f(x) = sin (x/2), then there exists a number c in the interval pi/2 < x < (3*pi / 2 that satisfies the conclusion of the Mean Value Theorem. Which of teh following could be c?
Here's my work...
The mean value theorem is f(c) = 1/(b-a) * int (f(x), x, a, b).
the indefinate integral of sin (x/2) is -2 cos (x/2) + c
c = -2 (cos ((3 * pi / 2) / 2) - cos ((pi / 2) / 2) / pi
c = -2 (cos (3 * pi / 4) - cos (pi / 4)) / pi
c = -2 ( - sqrt (2) / 2 - sqrt (2) / 2) / pi
c = -2 (- 2 sqrt (2) / 2) / pi
c = -2 (- sqrt (2)) / pi
c = 2 * sqrt (2) / pi
but, the answer sheet only gives answers with pi in the numerator (so i know i must be wrong). Can anyone help?