Chapter 5.5, Problem 62E

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

To determine

To calculate: The total basal metabolism rate of a young man.

Answer to Problem 62E

The total basal metabolism rate of a young man is $\underset{\_}{2040\text{kcal}}$.

Explanation of Solution

Given:

The function is $R\left(t\right)=85-0.18\cos \left(\frac{\pi t}{12}\right)$.

Show the integral function is,

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

The region lies between $t=0$ and $t=24$.

Calculation:

Take the consideration as follows:

$u=\frac{\pi t}{12}$ (1)

Differentiate both sides of the Equation (1).

$\begin{array}{l}du=\frac{\pi }{12}dt\\ dt=\frac{12}{\pi }du\end{array}$

Calculate the lower limit value of u using Equation (1).

Substitute 0 for t in Equation (1).

$\begin{array}{c}u=\frac{\pi \left(0\right)}{12}\\ =0\end{array}$

Calculate the upper limit value of u using Equation (1).

Substitute 24 for t in Equation (1).

$\begin{array}{c}u=\frac{\pi \left(24\right)}{12}\\ =2\pi \end{array}$

The integral function is,

${\displaystyle {\int }_0^{24}\left[85-0.18\cos \left(\frac{\pi t}{12}\right)\right]}dt$ (2)

Apply lower and upper limits for u in Equation (2).

Substitute u for $\left(\frac{\pi t}{12}\right)$ and $\left(\frac{12}{\pi }du\right)$ for dt in Equation (2).

$\begin{array}{c}{\displaystyle {\int }_0^{24}\left[85-0.18\cos \left(\frac{\pi t}{12}\right)\right]}dt={\displaystyle {\int }_0^{2\pi }\left(85-0.18\cos u\right)}\left(\frac{12}{\pi }du\right)\\ =\frac{12}{\pi }{\displaystyle {\int }_0^{2\pi }\left(85-0.18\cos u\right)}du\end{array}$ (3)

Integrate Equation (3).

$\begin{array}{c}\frac{12}{\pi }{\displaystyle {\int }_0^{2\pi }\left(85-0.18\cos u\right)}du=\frac{12}{\pi }{\left[85u-0.18\sin u\right]}_0^{2\pi }\\ =\frac{12}{\pi }\left\{\left[85\left(2\pi \right)-0.18\sin \left(2\pi \right)\right]-\left[85\left(0\right)-0.18\sin \left(0\right)\right]\right\}\\ =\frac{12}{\pi }\left[\left(170\pi -0\right)-\left(0-0\right)\right]\\ =\frac{12}{\pi }\times 170\pi \end{array}$

$=2040$

Hence, the total basal metabolism rate of the young man is $\underset{\_}{2040\text{kcal}}$.