In a certain city the temperature (in degrees Fahrenheit) t hours after 9 a.m. was approximated by the function
T(t) = 50 + 14sin((pi*t)/12).
Find the average temperature (in degrees Fahrenheit) during the period from 9 a.m. to 9 p.m.
I got (1/12)((12/pi)(50(12)-14cos((pi)))+(12/pi)(50(0)-14cos((0)/12))) but it's wrong...
Use wolframalpha website, it helps!
Thank you for your time!
1/12-0 integral from 0 to 12 for 50+14 sin(pit/12)
1/12 [50 t-14.12/pi cos pit/12] from 0 to 12
50 t28/pi
=58.9
To find the average temperature during the period from 9 a.m. to 9 p.m., we need to find and evaluate the definite integral of the temperature function T(t) from 9 a.m. (t = 0) to 9 p.m. (t = 12).
The integral of T(t) with respect to t is as follows:
∫(50 + 14sin((πt)/12)) dt
To evaluate this integral, we first integrate the constant term 50, which gives us 50t. Then we integrate the sine term using the substitution u = (πt)/12. The derivative of u with respect to t is du/dt = π/12, and thus dt = (12/π) du. The integral becomes:
50t - (12/π) ∫(14sin(u)) du
We can simplify the integral of sin(u) as follows:
∫sin(u) du = -cos(u) + C
where C is the constant of integration. Plugging this back into our integral expression, we get:
50t - (12/π)(-14cos(u) + C)
Now, we need to evaluate this expression between the limits of 0 to 12 (9 a.m. to 9 p.m.). Plugging in these limits for t and u, we have:
Average temperature = [50(12) - (12/π)(-14cos((π*12)/12) - (12/π)(-14cos((π*0)/12)]
Simplifying further, we have:
Average temperature = [600 + (12/π)(14cos(π) - 14cos(0)]
Since cos(π) = -1 and cos(0) = 1, we can simplify this expression to:
Average temperature = [600 + (12/π)(-14 - (-14))]
Average temperature = [600 + (12/π)(0)]
Average temperature = 600
Therefore, the average temperature during the period from 9 a.m. to 9 p.m. is 600 degrees Fahrenheit.
Apologies for the confusion, but the previous expression you mentioned gives the correct result. It seems there might have been a calculation error when you evaluated it. If you're still having trouble, using an online tool like Wolfram Alpha can provide a reliable calculation.