Possible Integration Problem

[cmajor47]

Homework Statement

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).

Homework Equations

T=61+18cos(πt/6)

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.

 

[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?

 

[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].

 

[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?

 

[rootX]

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?

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

 

[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.

 

[HallsofIvy]

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

I tried using the Mean Value Theorem and came up with this.
$$\frac{1}{5-2}\int_2^{5} 61+18\cos\frac{\pi t}{6} dt = 61+18\cos\frac{\pi t_{0}}{6}$$
No, that is NOT correct! For one thing the integral of 61 is 61 t, for another you haven't yet evaluated at the limits of integration and finally, you have dropped the "1/(5-2)".

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

Okay, now you have put them in but do you see that you shouldn't write "=" in the first row when they are NOT equal?

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

$$t_{0}\approx 3.449$$

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