Chapter 6.2, Problem 31E

To determine

To express: The given quantity in definite integral also to evaluate the integral using Theorem 2.

Answer to Problem 31E

The given quantity in integral is ${\displaystyle {\int }_6^{7.5}300dt}$

The value of integral is 450 calories

Explanation of Solution

Given Information:

A walker burns 300 calories per hour between 6:00 P.M and 7:30 P.M

Formula used:

Here, Theorem 2 is used to evaluate the given integral. Theorem 2 gives the integral value for a constant as,

  $\begin{array}{c}{\displaystyle {\int }_a^{b}sdt}=s{\left(t\right)}_a^b\\ =s(b-a)\end{array}$

Calculation Here, the walker burns calories at the rate of 300 calories /hour. So for the given range of time [6:00 PM _, 7:30 PM], the total calories burned can be represented using the integral, ${\displaystyle {\int }_6^{7.5}300dt}$ .

Using Theorem 2, the above integral can be solved as,

  $\begin{array}{c}{\displaystyle {\int }_6^{7.5}300dt}=300{\left(t\right)}_6^{7.5}\\ =300\left(7.5-6\right)\\ =300\times 1.5\\ =450\end{array}$

Thus, 450 calories are burned in the duration [6:00 PM _, 7:30 PM].