# Bad question on a take home quiz

1. Apr 1, 2004

### 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?

2. Apr 1, 2004

### matt grime

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

3. Apr 1, 2004

### 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...

4. Apr 1, 2004

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?

5. Apr 2, 2004

### 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.

6. Apr 3, 2004

### 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.

7. Apr 3, 2004

### matt grime

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

8. Apr 3, 2004

### 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.

9. Apr 3, 2004

### HallsofIvy

Staff Emeritus
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.

10. Apr 3, 2004

### 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?

11. Apr 3, 2004

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

12. Apr 3, 2004

### 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)

13. Apr 3, 2004

Ah, you're right. Sorry.

Your error is in f'(c).

14. Apr 3, 2004

### 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.

15. Apr 3, 2004

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

16. Apr 4, 2004

### 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

17. Apr 4, 2004

### 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.