# How do you use the Trapezoidal rule and three subintervals to give an estimate for the area between y=cscx and the x-axis from x= pi/8 to x = 7pi/8?

Aug 11, 2017

Approximately $3.75$ square units

#### Explanation:

Dividing the given interval $\left[\frac{\pi}{8} , \frac{7 \pi}{8}\right]$
into 3 equal width intervals (each with a width of $\frac{\pi}{4}$:
$\textcolor{w h i t e}{\text{XXX}} {I}_{1} = \left[\frac{\pi}{8} , \frac{3 \pi}{8}\right]$
$\textcolor{w h i t e}{\text{XXX}} {I}_{2} = \left[\frac{3 \pi}{8} , \frac{5 \pi}{8}\right]$
$\textcolor{w h i t e}{\text{XXX}} {i}_{3} = \left[\frac{5 \pi}{8} , \frac{7 \pi}{8}\right]$

[from this point on, extensive use of spreadsheet/calculator is recommended]

Evaluating $\csc \left(x\right)$ at each interval edge:
$\textcolor{w h i t e}{\text{XXX}} \csc \left(\frac{\pi}{8}\right) = 2.6131259298$
$\textcolor{w h i t e}{\text{XXX}} \csc \left(\frac{3 \pi}{8}\right) = 1.0823922003$
$\textcolor{w h i t e}{\text{XXX}} \csc \left(\frac{5 \pi}{8}\right) = 1.0823922003$
$\textcolor{w h i t e}{\text{XXX}} \csc \left(\frac{7 \pi}{8}\right) = 2.6131259298$

The area of each interval trapezoid is calculated as
$\textcolor{w h i t e}{\text{XXX}}$the average of the 2 edge lengths times the strip width.

${\text{Area}}_{I 1} = \frac{2.6131259298 + 1.0823922003}{2} \times \frac{\pi}{4}$
$\textcolor{w h i t e}{\text{XXXX}} = 1.4512265761$

Similarly we can calculate
${\text{Area}}_{I 2} = 0.8501088462$

${\text{Area}}_{I 3} = 1.4512265761$

and the Sum of these Areas gives an approximation of the integral value:
${\int}_{\frac{\pi}{8}}^{\frac{7 \pi}{8}} \csc \left(x\right) \mathrm{dx} \approx 3.7525619983$ Aug 11, 2017

${\int}_{\frac{\pi}{8}}^{\frac{7 \pi}{8}} \setminus \csc x \setminus \mathrm{dx} \approx 3.7526 \setminus \setminus \left(4 \mathrm{dp}\right)$

#### Explanation:

We have:

$y = \csc x$

We want to estimate $\int \setminus y \setminus \mathrm{dx}$ over the interval $\left[\frac{\pi}{8} , \frac{7 \pi}{8}\right]$ with $3$ strips; thus:

$\Delta x = \frac{\frac{7 \pi}{8} - \frac{\pi}{8}}{3} = \frac{\pi}{4}$

The values of the function, working to 6dp, are tabulated using Excel as follows; Trapezium Rule

$A = {\int}_{\frac{\pi}{8}}^{\frac{7 \pi}{8}} \setminus \csc x \setminus \mathrm{dx}$
$\setminus \setminus \setminus \approx \frac{0.785398}{2} \cdot \left\{2.613126 + 2.613126 + 2 \cdot \left(1.082392 + 1.082392\right)\right\}$
$\setminus \setminus \setminus = 0.392699 \cdot \left\{5.226252 + 2 \cdot \left(2.164784\right)\right\}$
$\setminus \setminus \setminus = 0.392699 \cdot \left\{5.226252 + 4.329569\right\}$
$\setminus \setminus \setminus = 0.392699 \cdot 9.555821$
$\setminus \setminus \setminus = 3.752562$

Actual Value

For comparison of accuracy:

$A = {\int}_{\frac{\pi}{8}}^{\frac{7 \pi}{8}} \setminus \csc x \setminus \mathrm{dx}$
 \ \ \ = [ color(white)(""/"") -log(abs(csc(x)+cot(x))) \ ]_(pi/8)^((7pi)/8
$\setminus \setminus \setminus = 3.229781832346191$