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#?

2 Answers
Aug 11, 2017

Approximately #3.75# square units

Explanation:

Dividing the given interval #[pi/8,(7pi)/8]#
into 3 equal width intervals (each with a width of #pi/4#:
#color(white)("XXX")I_1=[pi/8,(3pi)/8]#
#color(white)("XXX")I_2=[(3pi)/8,(5pi)/8]#
#color(white)("XXX")i_3=[(5pi)/8,(7pi)/8]#

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

Evaluating #csc(x)# at each interval edge:
#color(white)("XXX")csc(pi/8)=2.6131259298#
#color(white)("XXX")csc((3pi)/8)=1.0823922003 #
#color(white)("XXX")csc((5pi)/8)=1.0823922003 #
#color(white)("XXX")csc((7pi)/8)=2.6131259298#

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

#"Area"_(I1)=(2.6131259298+1.0823922003)/2 xx pi/4#
#color(white)("XXXX")=1.4512265761#

Similarly we can calculate
#"Area"_(I2)=0.8501088462 #

#"Area"_(I3)=1.4512265761#

and the Sum of these Areas gives an approximation of the integral value:
#int_(pi/8)^((7pi)/8)csc(x) dx~~3.7525619983#

enter image source here

Aug 11, 2017

# 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;

enter image source here

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 #