Chapter 8.EA, Problem 1EA

To determine

(a)

To find:

The approximate value of integral by using the trapezoidal rule for the given condition.

Answer to Problem 1EA

Solution:

The approximate value of integral by using the trapezoidal rule for the given condition is 46.

Explanation of Solution

Given:

The given condition is:

$F=\frac{V_1\left(T_b-T_i\right)}{{\displaystyle {\int }_0^{\infty }\Delta Tdt}}$

Approach:

Trapezoidal Rule:

Let f be a continuous function on $\left[a,b\right]$ and let $\left[a,b\right]$ be divided into n equal subintervals by the points $a=x_0,x_1,x_2,\cdots ,x_n=b$. Then, by the trapezoidal rule,

${\displaystyle {\int }_a^{b}f\left(x\right)dx\approx \left(\frac{b-a}{n}\right)\left[\frac{1}{2}f\left(x_0\right)+f\left(x_1\right)+\cdots +f\left(x_{n-1}\right)+\frac{1}{2}f\left(x_n\right)\right]}$.

Calculation:

Consider the given function,

$F=\frac{V_1\left(T_b-T_i\right)}{{\displaystyle {\int }_0^{\infty }\Delta Tdt}}$

We calculate the discharge during a 24-hour period.

$\begin{array}{rll}{\displaystyle {\int }_0^{24}c\left(t\right)dt}& =& {\displaystyle \sum _{i=1}^{6}c\left(t_i\right)\Delta t}\\ & =& \Delta t{\displaystyle \sum _{i=1}^{6}c\left(t_i\right)}\\ & =& 4\left(3.0+3.5+2.5+1.0+0.5+1.0\right)\\ & =& 46\end{array}$

Hence, the total daily discharge is:

$\begin{array}{rll}{Q}_m& =& F{\displaystyle {\int }_0^{24}c\left(t\right)dt}\\ & =& \left(1.02\times {10}^5\frac{{\text{m}}^3}{\text{hour}}\right)\left(46\frac{\text{mg}\times \text{hour}}{{\text{m}}^3}\right)\\ & =& 4.7\text{ kg}\end{array}$

Then, $x_{i+1}$ cab be calculated by $x_{i+1}=x_i+\frac{b-a}{n}$.

The corresponding function values will be calculated as below:

$x_i$	$f\left(x_i\right)$
0	3
4	3.5
8	2.5
12	1.0
16	5.0
20	1.0
24	3.0

Substitute the value in the formula, we get:

$\begin{array}{rll}{\displaystyle {\int }_0^{24}f\left(x\right)dx}& =& \left(\frac{24}{6}\right)\left[\frac{1}{2}\left(3\right)+3.5+2.5+1+5+1+\frac{1}{2}\left(3\right)\right]\\ & =& 46\end{array}$

Hence, the approximate value of integral by using the trapezoidal rule for the given condition is 46.

To determine

(b)

To find:

The approximate value of integral by using the Simpson’s rule for the given condition.

Answer to Problem 1EA

Solution:

The approximate value of integral by using the Simpson’s rule for the given condition is $45.333$.

Explanation of Solution

Given:

The given condition is:

$F=\frac{V_1\left(T_b-T_i\right)}{{\displaystyle {\int }_0^{\infty }\Delta Tdt}}$

Approach:

Simpson’s Rule:

Let f be a continuous function on $\left[a,b\right]$ and let $\left[a,b\right]$ be divided into n equal subintervals by the points $a=x_0,x_1,x_2,\cdots ,x_n=b$. Then, by the Simpson’s rule,

${\displaystyle {\int }_a^{b}f\left(x\right)dx\approx \left(\frac{b-a}{3n}\right)\left[f\left(x_0\right)+4f\left(x_1\right)+\cdots +4f\left(x_{n-1}\right)+f\left(x_n\right)\right]}$.

Calculation:

Consider the given function,

$F=\frac{V_1\left(T_b-T_i\right)}{{\displaystyle {\int }_0^{\infty }\Delta Tdt}}$

Then, $x_{i+1}$ cab be calculated by $x_{i+1}=x_i+\frac{b-a}{n}$.

The corresponding function values will be calculated as below:

$x_i$	$f\left(x_i\right)$
0	3
4	3.5
8	2.5
12	1.0
16	5.0
20	1.0
24	3.0

Substitute the value in the formula, we get:

$\begin{array}{rll}{\displaystyle {\int }_0^{24}f\left(x\right)dx}& =& \left(\frac{24}{6\cdot 3}\right)\left[\frac{1}{2}\left(3\right)+3.5+2.5+1+5+1+\frac{1}{2}\left(3\right)\right]\\ & =& \frac{4}{3}\left(34\right)\\ & =& 45.333\end{array}$

Hence, the approximate value of integral by using the Simpson’s rule for the given condition is $45.333$.