How do you use the Trapezoidal rule and three subintervals to give an estimate for the area between $y = csc x$ $y=cscx$ and the x-axis from $x = \frac{π}{8}$ $x= pi/8$ to $x = 7 \frac{π}{8}$ $x = 7pi/8$ ?

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How do you use the Trapezoidal rule and three subintervals to give an estimate for the area between $y = csc x$ $y=cscx$ and the x-axis from $x = \frac{π}{8}$ $x= pi/8$ to $x = 7 \frac{π}{8}$ $x = 7pi/8$ ?

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Answer 1 · 2017-08-11T20:23:29

Approximately $3.75$ $3.75$ square units

Explanation:

Dividing the given interval $[\frac{π}{8}, \frac{7 π}{8}]$ $[pi/8,(7pi)/8]$
into 3 equal width intervals (each with a width of $\frac{π}{4}$ $pi/4$ :
$XXX I_{1} = [\frac{π}{8}, \frac{3 π}{8}]$ $color(white)("XXX")I_1=[pi/8,(3pi)/8]$
$XXX I_{2} = [\frac{3 π}{8}, \frac{5 π}{8}]$ $color(white)("XXX")I_2=[(3pi)/8,(5pi)/8]$
$XXX i_{3} = [\frac{5 π}{8}, \frac{7 π}{8}]$ $color(white)("XXX")i_3=[(5pi)/8,(7pi)/8]$

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

Evaluating $csc (x)$ $csc(x)$ at each interval edge:
$XXX csc (\frac{π}{8}) = 2.6131259298$ $color(white)("XXX")csc(pi/8)=2.6131259298$
$XXX csc (\frac{3 π}{8}) = 1.0823922003$ $color(white)("XXX")csc((3pi)/8)=1.0823922003$
$XXX csc (\frac{5 π}{8}) = 1.0823922003$ $color(white)("XXX")csc((5pi)/8)=1.0823922003$
$XXX csc (\frac{7 π}{8}) = 2.6131259298$ $color(white)("XXX")csc((7pi)/8)=2.6131259298$

The area of each interval trapezoid is calculated as
$XXX$ $color(white)("XXX")$ the average of the 2 edge lengths times the strip width.

${Area}_{I 1} = \frac{2.6131259298 + 1.0823922003}{2} \times \frac{π}{4}$ $"Area"_(I1)=(2.6131259298+1.0823922003)/2 xx pi/4$
$XXXX = 1.4512265761$ $color(white)("XXXX")=1.4512265761$

Similarly we can calculate
${Area}_{I 2} = 0.8501088462$ $"Area"_(I2)=0.8501088462$

${Area}_{I 3} = 1.4512265761$ $"Area"_(I3)=1.4512265761$

and the Sum of these Areas gives an approximation of the integral value:
$\int_{\frac{π}{8}}^{\frac{7 π}{8}} csc (x) d x \approx 3.7525619983$ $int_(pi/8)^((7pi)/8)csc(x) dx~~3.7525619983$

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Answer 2 · 2017-08-11T21:32:41

$int_(pi/8)^((7pi)/8) \ cscx \ dx ~~ 3.7526 \ \ (4dp)$

Explanation:

We have:

$y = cscx$

We want to estimate $int \ y \ dx$ over the interval $[pi/8,(7pi)/8]$ with $3$ strips; thus:

$Deltax = ((7pi)/8-pi/8)/3 = pi/4$

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

Trapezium Rule

$A = int_(pi/8)^((7pi)/8) \ cscx \ dx$
$\ \ \ ~~ 0.785398/2 * { 2.613126 + 2.613126 + 2*(1.082392 + 1.082392) }$
$\ \ \ = 0.392699 * { 5.226252 + 2*(2.164784) }$
$\ \ \ = 0.392699 * { 5.226252 + 4.329569 }$
$\ \ \ = 0.392699 * 9.555821$
$\ \ \ = 3.752562$

Actual Value

For comparison of accuracy:

$A= int_(pi/8)^((7pi)/8) \ cscx \ dx$
$\ \ \ = [ color(white)(""/"") -log(abs(csc(x)+cot(x))) \ ]_(pi/8)^((7pi)/8$
$\ \ \ = 3.229781832346191$

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How do you use the Trapezoidal rule and three subintervals to give an estimate for the area between $y = csc x$ $y=cscx$ and the x-axis from $x = \frac{π}{8}$ $x= pi/8$ to $x = 7 \frac{π}{8}$ $x = 7pi/8$ ?

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How do you use the Trapezoidal rule and three subintervals to give an estimate for the area between y=cscxy=cscxy=cscx and the x-axis from x= pi/8x=π8x= pi/8 to x = 7pi/8x=7π8x = 7pi/8?

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Explanation:

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How do you use the Trapezoidal rule and three subintervals to give an estimate for the area between $y = csc x$ $y=cscx$ and the x-axis from $x = \frac{π}{8}$ $x= pi/8$ to $x = 7 \frac{π}{8}$ $x = 7pi/8$ ?