A model for the basal metabolism rate in kcal/h, of a young man is R(t)=85-0.18cos(pi(t)/12), where t is the time in hours measured from 5:00 am. What is the total basal metabolism of this man, over a 24-hour time period?

Question

mony01 · Answer

$$\int\limits_{0}^{24}R(t) dt$$

ganeshie8 · Answer

looks good.
the phrase 'time is measured from 5:00am' is begginhog us to set bounds as 5->29
but u wil get the same answer so it should be okay

mony01 · Answer

so the integral is $$\int\limits_{0}^{24}85-0.18\cos(\frac{ \pi(t) }{12 })$$

ganeshie8 · Answer

yes, but i prefer setting it up as :   

$ \int\limits_{5}^{29}85-0.18\cos(\frac{ \pi(t) }{12 }) dt $

mony01 · Answer

is it 5 to 24?

ganeshie8 · Answer

5 to 29

mony01 · Answer

what would the answer be?

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=int_5%5E29+%2885-0.18cos%28pi*t%2F12%29%29+dt

mony01 · Answer

can you tell me if this is the set up when you take the anti-derivative 
85t-0.18sin(pi(t)/12)^2/2 dt

ganeshie8 · Answer

$\large \int\limits_{5}^{29}85-0.18\cos(\frac{ \pi(t) }{12 }) dt$

$\large  \int\limits_{5}^{29}85 dt -\int\limits_{5}^{29} 0.18\cos(\frac{ \pi(t) }{12 }) dt$

$\large  85t \Big|_5^{29} -\frac{0.18\sin(\frac{ \pi(t) }{12 })}{\frac{\pi}{12}} \Big|_5^{29}$

ganeshie8 · Answer

take the limits now