Chapter 6.2, Problem 29E

To determine

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

Answer to Problem 29E

The given quantity in integral is ${\displaystyle {\int }_8^{11}87dt}$

The value of integral is 261 m

Explanation of Solution

Given Information:

The train is moving at a speed of 87 mph from 8:00 A.M to 11.00 A.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 The distance traveled by the train moving at a constant speed of s for a given range of time [ t_{1 ,} t₂ ] is given by the integral, ${\displaystyle {\int }_{t_1}^{t_2}sdt}$ .

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

  ${{\displaystyle {\int }_{t_1}^{t_2}sdt=s\left[t\right]}}_{t_1}^{t_2}=s(t_2-t_1)$

Here, speed is 87 mph for a constant time duration which is between 8:00 A.M to 11.00 A.M

Considering the given integral and speed, the distance can be calculated as

  $\begin{array}{c}{{\displaystyle {\int }_8^{11}87dt=87\left[t\right]}}_8^{11}\\ =87\times \left(11-8\right)\\ =261\text{ m}\end{array}$