In Seattle on September 30, the temperature hours after midnight was given by the function y=60 + 12sin((pi/x)(x-11))

What was the average temperature over the period from 8 A.M. until 10 P.M.?

The function does not exist at midnight (x=0)and is "weird" between 1am and 7am, as the temperature goes through many cycles. Eventually, it rises as the day goes on until mid-night.

In any case, to find the average from 8am to 10pm, we would integrate
f(x)=60 + 12sin((pi/x)(x-11))
with respect to x from x=8 to x=22, and divide by the time difference (22-8)=14.
The integral, however, is not easy to calculate because it involves gamma function.
If you have done numerical integration before (such as simpson's rule,etc.) you can get the answer using numerical integration to get 65.8 degrees.

Well, it sounds like you're trying to calculate the average temperature in Seattle over a specific time period. But let's face it, talking about averages can be as exciting as watching paint dry. So instead, let's try to spruce things up a bit with a little joke!

Why did the temperature go to therapy?

Because it had too many ups and downs!

Now, let's get back to your question. To find the average temperature from 8 A.M. until 10 P.M., we need to calculate the total temperature during that time period and then divide it by the number of hours.

Since there are 12 hours between 8 A.M. and 10 P.M., we can evaluate the given function for each hour and sum up the results. After that, we'll divide the total by 12 to get the average temperature.

Does that make sense?

To find the average temperature over the period from 8 A.M. until 10 P.M., we need to find the average value of the temperature function y = 60 + 12sin((π/x)(x-11)) over that time interval.

To do this, we'll integrate the function over the interval from 8 to 22 (representing 8 A.M. to 10 P.M.) and then divide the result by the length of the interval (14 hours).

∫[8,22] (60 + 12sin((π/x)(x-11))) dx / 14

Now, let's find the integral:

∫[8,22] (60 + 12sin((π/x)(x-11))) dx / 14
= [60x + 12∫sin((π/x)(x-11)) dx] / 14 [using linearity of integration]

The integral of sin function is -cos, so we have:
= [60x - 12cos((π/x)(x-11))] / 14 + C

Now, we need to evaluate this expression at the upper and lower limits of integration:

[60(22) - 12cos((π/22)(22-11))] / 14 - [60(8) - 12cos((π/8)(8-11))] / 14 + C
= (1320 - 12cos(π)) / 14 - (480 - 12cos(3π/8)) / 14 + C [cos(π) = -1 and cos(3π/8) = (sqrt(2 + sqrt(2)) / 2)]

Simplifying further:
= (1320 + 12) / 14 - (480 - 12(sqrt(2 + sqrt(2)))) / 14 + C
= (1332 / 14) - (480 - 12(sqrt(2 + sqrt(2)))) / 14 + C
= 95.14 - (480 - 12(sqrt(2 + sqrt(2)))) / 14 + C

Therefore, the average temperature over the period from 8 A.M. until 10 P.M. in Seattle on September 30 is approximately 95.14 - (480 - 12(sqrt(2 + sqrt(2)))) / 14 + C degrees Fahrenheit.

To find the average temperature over the period from 8 A.M. until 10 P.M., we need to calculate the definite integral of the given function over that time period and then divide it by the number of hours in that time period.

The time period from 8 A.M. until 10 P.M. is 14 hours.

So, we need to calculate the definite integral of the function y=60 + 12sin((pi/x)(x-11)) from 8 to 22 (representing the 24-hour clock).

To integrate the function, we can split it into two parts: the constant term (60) and the sine term.

Integrating the constant term:
∫ 60 dx = 60x

Integrating the sine term requires a u-substitution. Let u = (π/x)(x-11).

Substituting and differentiating:
du/dx = π(x-11)(d/dx)(1/x) - (π/x)(d/dx)(x-11)
du/dx = -π(x-11)/x^2

Rearranging, we get:
dx = -x^2/(π(x-11)) du

Now, substituting in the integral:
∫ 12sin((π/x)(x-11)) dx = -12∫ sin(u) (-x^2/(π(x-11))) du
∫ 12sin((π/x)(x-11)) dx = 12∫ x^2/(π(x-11)) sin(u) du

Simplifying the integral and taking out the constant:
= -12/π ∫ x/(x-11) sin(u) du

Now to solve this subintegral, we can use integration by parts with u = sin(u) and v' = x/(x-11).

Using the integration by parts formula:
∫ u dv = uv - ∫ v du

Let's calculate each part separately:

u = sin(u)
du = cos(u) du

v = ∫ x/(x-11) dx
v = ∫ (x-11+11)/(x-11) dx
v = ∫ (1 + 11/(x-11)) dx
v = x + 11 ln| x - 11|

Now, let's substitute the values into the formula:

-12/π ∫ x/(x-11) sin(u) du
= -12/π (x + 11 ln| x - 11|)(sin(u)) + 12/π ∫ (x + 11 ln| x - 11|) cos(u) du

Now, we can integrate the second part using the integration by parts again.

u = x + 11 ln| x - 11|
du = 1 + 11/(x-11) dx

v = ∫ cos(u) du
v = sin(u)

Plugging in the values, we get:

-12/π (x + 11 ln| x - 11|)(sin(u)) + 12/π [sin(u)(x + 11 ln| x - 11|) - ∫ (1 + 11/(x-11))(sin(u)) dx]

Now we have the final expression for the definite integral from 8 to 22:

-12/π [x + 11 ln| x - 11|)(sin(u)) + 12/π [sin(u)(x + 11 ln| x - 11|) - ∫ (1 + 11/(x-11))(sin(u)) dx]] from 8 to 22

Now, we can evaluate the definite integral by substituting the limits of integration (8 and 22) into this expression, and then divide the result by the number of hours (14) to get the average temperature over the period from 8 A.M. until 10 P.M.