Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 56

What is the total basal metabolism of a man, $\displaystyle \int^{24}_0 R(t) dt$, over a 24-hour time period? Suppose that a model for the basal metabolism rate, in kcal/h, of a man is $\displaystyle R(t) = 85 - 0.18 \cos \left( \frac{\pi t}{12} \right)$, where $t$ is the time in hours measured from 5:00 am.

$\displaystyle \int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt$

Let $\displaystyle u = \frac{\pi t}{12} dt$, then $\displaystyle du = \frac{\pi}{12} dt$, so $\displaystyle dt = \frac{12du}{\pi}$. When $t = 0, u = 0$ and when $t = 24, u = 2 \pi$. Thus,



$
\begin{equation}
\begin{aligned}

\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& \int^{24}_0 85 - 0.18 \cos u \frac{12 du}{\pi}
\\
\\
\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& \frac{12}{\pi} \int^{24}_0 85 - 0.18 \cos u du
\\
\\
\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& \frac{12}{\pi} \int^{24}_0 85 du - \frac{2.16}{\pi} \int^{24}_0 \cos u du
\\
\\
\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& \left. \frac{12}{\pi} \cdot 85 u \right|^{24}_0 - \left. \frac{2.16}{\pi} \cdot \sin u \right|^{24}_0
\\
\\
\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& \frac{12 }{\pi} [85(2 \pi) - 85 (0)] - \frac{2.16}{\pi} [\sin (2 \pi) - \sin (0)]
\\
\\
\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& \frac{12}{\cancel{\pi}} (170 \cancel{\pi}) - \frac{2.16}{\pi} (0)
\\
\\
\int^{24}_0 85 - 0.18 \cos \left( \frac{\pi t}{12} \right) dt =& 2040 \text{kcal/h}


\end{aligned}
\end{equation}
$