Question

a model for the basal metabolism rate, in kcal/h, of a young man is ( r(t)=80 - 0.16cosleft(\frac{pi t}{12}
ight) ), where ( t ) is the time in hours measured from 5:00 a.m. what is the total basal metabolism of this man, ( int_{0}^{24}r(t)dt ), over a 24 - hour period?

kcal

Explanation:

Step1: Integrate the function

We know that \(\int_{0}^{24}(80 - 0.16\cos(\frac{\pi t}{12}))dt=\int_{0}^{24}80dt-\int_{0}^{24}0.16\cos(\frac{\pi t}{12})dt\).
For \(\int_{0}^{24}80dt\), using the formula \(\int_{a}^{b}kdt=k(b - a)\) (\(k = 80,a = 0,b = 24\)), we have \(80\times(24 - 0)=1920\).
For \(\int_{0}^{24}0.16\cos(\frac{\pi t}{12})dt\), let \(u=\frac{\pi t}{12}\), then \(dt=\frac{12}{\pi}du\). When \(t = 0\), \(u = 0\); when \(t = 24\), \(u = 2\pi\).
So \(\int_{0}^{24}0.16\cos(\frac{\pi t}{12})dt=0.16\times\frac{12}{\pi}\int_{0}^{2\pi}\cos(u)du\).
Since \(\int\cos(u)du=\sin(u)+C\), then \(0.16\times\frac{12}{\pi}[\sin(u)]_{0}^{2\pi}=0.16\times\frac{12}{\pi}(\sin(2\pi)-\sin(0)) = 0\).

Step2: Calculate the result

\(\int_{0}^{24}(80 - 0.16\cos(\frac{\pi t}{12}))dt=1920-0\).

Answer:

\(1920\)