bad question on a take home quiz

[DavidWi]

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?

[matt grime]

You've dropped the f from f(c), as well as used c for two different things (the constant of the integration).

[DavidWi]

As far as I know, the constant of integration only applies to taking the indefinate integral. Since I'm taking the definate integral, the constant of integration cancels out. Expalin the rest though...

[cookiemonster]

Your first equation was:

f(c) = \frac{\int_a^b f(x) \, dx}{b-a}

Notice the left-hand side of this equation.

Now look at the left-hand side of all your work. See something missing?

cookiemonster

[matt grime]

Using c for two things was just a note on being careful with your notation. And you spell definite with is not an a.

[DavidWi]

soooo... what you guys are saying is that I should set (2 * sqrt (2)) / pi equal to my function and solve for x...

2 * sqrt (2) / pi = sin (x / 2)
x / 2 = arcsin (2 * sqrt (2) / pi)
x = 2 * arcsin (2 * sqrt (2) / pi)

But this still isn't any of my answers... I'm still doing something wrong. I dont' know what it is.

[matt grime]

You say which of the choices might be c in the original post. Is there one that is closest?

[DavidWi]

The five answers are (A) 2 pi / 3, (B) 3 pi / 4, (C) 5 pi / 6 (D) pi, (E) 3 pi / 2

I can't figure out a way to simplify arcsin (2 sqrt (2) / pi), so I don't think it's any of these answers.

[HallsofIvy]

The correct answer is 2arcsin(2&radic;(2)/pi) which lies between A and B.

That's assuming you mean the "integral" mean value theorem. Normally when I see just "mean value theorem", I think "differential" mean value theorem. The value of c for that is quite easily pi.

[DavidWi]

ohhhh, you're right! wow! I guess I got too caught up in the integral mean value theorum. But there's the much easier to do differential one.

f'(c) = (sin (3 pi / 4) - sin (pi / 4)) / pi
f'(c) = (sqrt (2) / 2 - sqrt (2) / 2) / pi
f'(c) = 0 / pi = 0

sin c / 2 = 0
c / 2 = arcsin 0
c / 2 = 0
c = 0

Huh, still, I get the wrong answer. Any more help?

[cookiemonster]

In the first step, don't you mean sin(3pi/2), not sin(3pi/4)?

cookiemonster

[DavidWi]

no, i meant sin (3 pi / 4) because it's sin (x / 2).

if x = 3 pi / 2, then it would be sin (3 pi / 2 / 2) = sin (3 pi / 4)

[cookiemonster]

Ah, you're right. Sorry.

Your error is in f'(c).

cookiemonster

[DavidWi]

OHHHHH! I GOT IT! Maybe.... Maybe... I forgot to take into account my domain. I'm going from pi / 2 to 3 pi / 2. But when i solved for the arcsin 0, i thought it was a 0 radians, which is out of my domain.

arcsin 0 = pi, not 0

sin (c / 2) = 0
c / 2 = arcsin 0
c / 2 = pi
c = 2 * pi....

noooooo the stupid rotten 2... i thought i almost had it.

[cookiemonster]

f'(c)... As in d/dx(f(x))|c... or maybe d/dx(sin(x/2))|c...

cookiemonster

[himanshu121]

Mean value theorem

f'(c)= \frac{f( \frac{3 \pi}{2}) - f( \frac{ \pi}{2})}{ \pi}

u can see that

cos(c/2)= 0

therefore
c = pi

[DavidWi]

wow, finally i see the right answer. Now, I wonder if I can get my calc teacher with that problem. I don't think he'll be able to do it. Maybe he will though.