Chapter 6.2, Problem 32E

To determine

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

Answer to Problem 32E

The given quantity in integral is ${\displaystyle {\int }_{8.5}^{11}0.4dt}$

The value of integral is 1 liter

Explanation of Solution

Given Information:

Water leaks from a bucket at a rate of 0.4 liters per hour between 8:30 A.M and 11 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 Here, water leaks at the rate of 0.4 liters/hour. So for the given range of time [8:30 A.M_,11 A.M], the total water leak can be represented using the integral, ${\displaystyle {\int }_{8.5}^{11}0.4dt}$ .

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

  $\begin{array}{c}{\displaystyle {\int }_{8.5}^{11}0.4dt}=0.4{\left(t\right)}_{8.5}^{11}\\ =0.4\left(11-8.5\right)\\ =0.4\times 2.5\\ =1\end{array}$

Thus, 1 liter of water leaks from the bucket in the duration [8:30 A.M_,11 A.M] .