# Homework Help: Possible Integration Problem

1. Aug 18, 2008

### cmajor47

1. The problem statement, all variables and given/known data
The mean daily temperature in degrees fahrenheit in Athens, GA, t months after July 15 is closely approximated by T=61+18cos(πt/6)
Find the average temp. between September 15 (t=2) and December 25 (t=5).

2. Relevant equations
T=61+18cos(πt/6)

3. The attempt at a solution
Can I just plug in t=2 and t=5 into the equation, add the answers, then divide by two to get the average temp. Or do I have to use integration. I'm not sure because the question is asked in the chapter on integration but it doesn't look like I need to use it.

2. Aug 18, 2008

### CompuChip

I think you have to find the average of the function by integration. Adding and dividing by two works if the function is linear. For example, suppose that between t = 2 and t = 4.9 the temperature is 40 and between t = 4.9 and t = 5 it is -10. Then surely you would disagree if I said that the average temperature between t = 2 and t = 5 was (40 - 10)/2 = 15 degrees?

Last edited: Aug 18, 2008
3. Aug 18, 2008

### konthelion

You have to use integration. The reason for this is because your function $$t: [a,b] \rightarrow R$$ is continuous. By the Mean Value Theorem for Integrals then there is a point $$x_{0}$$ in [a,b] such that

$$\frac{1}{b-a} \int_a^{b} t = t(x_{0})$$ where [a,b] is [2,5].

4. Aug 18, 2008

### cmajor47

I tried using the Mean Value Theorem and came up with this.
$$\frac{1}{5-2}\int_2^{5} 61+18\cos\frac{\pi t}{6} dt = 61+18\cos\frac{\pi t_{0}}{6}$$

$$\frac{1}{3} 61t+\frac{108}{\pi}sin\frac{\pi t}{6} \mid_2^{5} = \frac{-\sqrt3+1}{\pi}$$

$$61+18cos\frac{\pi t_{0}}{6} = \frac{-\sqrt3+1}{\pi}$$

$$t_{0}\approx 3.449$$

When I plug this into $$61+18cos\frac{\pi t}{6}$$ I get approximately 56.8056. Is this right?

Last edited: Aug 18, 2008
5. Aug 18, 2008

### rootX

assuming that you got initial equations correct, I also got same answer ..

>> int ('1/3*(61+18*cos(pi*t/6))','t',2,5)

ans =

-(-61*pi+18*3^(1/2)-18)/pi

>> eval(ans)

ans =

56.8057

>> (ans-61)/18

ans =

-0.2330

>> acos(ans)

ans =

1.8060

>> ans*6/pi

ans =

3.4492

>>

EDIT: ONLY THING I WOULD BE WORRIED ABOUT IS USING ACOS VALUE OF CALCULATOR, IT IS WRONG SOMETIMES ... (I will see into this) but 3.4492 looks reasonable

Last edited: Aug 18, 2008
6. Aug 19, 2008

### CompuChip

The answer is correct. No need for the mean value theorem:
$$\operatorname{avg}\limits_{[a,b]} f(x) = \frac{1}{b - a} \int_a^b f(x) \, \mathrm dx$$
so in this case the average temperature is
$$61 + \frac{1}{3} \int 18 \cos(\pi t / 6) \, \mathrm dt = 61 + 6 \left[ \frac{6}{\pi} \sin(\pi t / 6) \right]_{t = 2}^{t = 5} = 61 + \frac{18}{\pi} \left( 1 - \sqrt{3} \right)$$
which evaluates numerically to approximately 56.8057.

7. Aug 19, 2008

### HallsofIvy

I must say that I am not at all crazy about the way you write things!