How do you find the area using the midpoint approximation method, given # sinx(dx) #, on the interval [0, pi] with n=10?

1 Answer
Jim H
Nov 20, 2016

I will use what I think is the usual notation throughout this solution.

#int_0^pi sinx dx#

Note that #f(x) = sinx# and #a=0# and #b=pi#

#n=10# So #Deltax = (b-a)/n = (pi-0)/10 =pi/10#

All endpoints: start with #a# and add #Deltax# successively:

#0# #underbrace(color(white)"XX")_(+pi/10)# #pi/10# #underbrace(color(white)"XX")_(+pi/10)# #(2pi)/10# #underbrace(color(white)"XX")_(+pi/10)# #(3pi)/10# . . . #(9pi)/10##underbrace(color(white)"XX")_(+pi/10)# #pi#

All endpoints: #pi/10#, #(2pi)/10#, #(3pi)/10#, . . . #(9pi)/10#, #pi#

The subintervals are:

#(0,pi/10)#, #(pi/10,(2pi)/10)#, #((2pi)/10, (3pi)/10)#, . . . #((9pi)/10,pi)#

We have been aske to use the midpoint of each subinterval as its sample point. The midpoints may be found by averaging the endpoints of each subinterval or by averaging the endpoints of the first subinterval to find its midpoint (#pi/20#) and then successively adding #Delta x# (#=(2pi)/20#) to get the others.

The midpoints are:

#pi/20#, #(3pi)/20#, #(5pi)/20#, . . . #(19pi)/20#

Now the Riemann sum is the sum of the area of the 4 rectangles. We find the area of each rectangle by

#"height" xx "base" = f("sample point") xx Deltax#

Here we are using midpoints for sample points. So

#R = f(pi/20) * pi/10+f((3pi)/20) * pi/10+f((5pi)/20) * pi/10+ * * * +f(pi) * pi/10#

#= (f(pi/20)+f((3pi)/20) +f((5pi)/20) + * * * +f(pi)) pi/10#

#= (sin(pi/20)+sin((3pi)/20) + sin((5pi)/20) + * * * +sin(pi)) pi/10#

Finish the arithmetic to finish.

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