Question

A model for the basal metabolism rate, in \({\rm{kcal/h}}\)of a young man is\({\rm{R(t) = 85 - 0}}{\rm{.18cos(\pi t/12)}}\), where \({\rm{t}}\)is the time in hours measured from 5:00 AM. What is the total?

Short answer

The total \({\rm{kcal/h}}\) is\({\rm{2040 kcal}}\).

Step-by-step solution

Given data.

\(\int_{\rm{0}}^{{\rm{24}}} {\rm{8}} {\rm{5 - 0}}{\rm{.18cos}}\left( {\frac{{{\rm{\pi t}}}}{{{\rm{12}}}}} \right){\rm{dt}}\)AF

Time 5:00 AM.

Changing Integration limits.

Let, \({\rm{u = }}\frac{{{\rm{\pi t}}}}{{{\rm{12}}}}\)

\({\rm{du = }}\frac{{\rm{\pi }}}{{{\rm{12}}}}{\rm{dt}}\)

And \({\rm{dt = }}\frac{{{\rm{12}}}}{{\rm{\pi }}}{\rm{du}}\)

Change the integration limits.

\({\rm{t = 0,u = }}\frac{{{\rm{\pi (0)}}}}{{{\rm{12}}}}{\rm{ = 0}}\) at \({\rm{t = 24,u = }}\frac{{{\rm{\pi (24)}}}}{{{\rm{12}}}}{\rm{ = 2\pi }}\).

Substituting the value.

\(\begin{aligned}{l}\frac{{{\rm{12}}}}{{\rm{\pi }}}\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{(85 - 0}}{\rm{.18cosu)}}} {\rm{du}}\\\left. {\frac{{{\rm{12}}}}{{\rm{\pi }}}{\rm{(85u - 0}}{\rm{.18sinu)}}} \right|_{\rm{0}}^{{\rm{2\pi }}}\\\frac{{{\rm{12}}}}{{\rm{\pi }}}{\rm{(85(2\pi ) - 0}}{\rm{.18sin(2\pi ) - 85(0) + 0}}{\rm{.18sin(0))}}\\\frac{{{\rm{12}}}}{{\rm{\pi }}}{\rm{(85(2\pi )) = 2040}}\end{aligned}\)

Therefore, the total \({\rm{kcal/h}}\) is\({\rm{2040 kcal}}\).