Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
What is the integral of sin(x) dx from 0 to 2pi?
An alternative way to do this starting from the limit definition is:
##int_(a)^(b) f(x)dx = lim_(n->oo) sum_(i=1)^N f(x_i^"*")Deltax##
where:
- ##n## is the number of rectangles used to approximate the integral, i.e. the area between the curve and the x-axis.
- ##i## is the index of each rectangle in ##[0,2pi]##.
- ##N## is the index of the final rectangle in ##[0,2pi]##.
- ##f(x_i^"*")## is the height of each given rectangle in ##[0,2pi]##, which varies as ##sin(x)##.
- ##Deltax## is the width of each given rectangle in ##[0,2pi]##, which converges to ##0## as ##n->oo##.
If we use the midpoint-rectangular approximation method (MRAM), we choose a convenient interval ##Deltax## such that we can find a midpoint for each rectangle of dimension ##Deltax xx f(x_i^"*")##, where the midpoint of the ##i##th rectangle is defined as
##M_i = x_(i-1)+(x_i - x_(i-1))/2##.
Let us choose ##Deltax = pi/2## such that ##x = {0,pi/2,pi,(3pi)/2,2pi}## for ##[0,2pi]## and ##n = {1,2,3,4}##.
Then each rectangle's width is ##pi/2##, and:
- Rectangle ##1## spans ##[0,pi/2]##.
- Rectangle ##2## spans ##[pi/2,pi]##.
- Rectangle ##3## spans ##[pi,(3pi)/2]##.
- Rectangle ##4## spans ##[(3pi)/2,2pi]##.
Each corresponds to an ##f(x_i^"*")## that gives you the height of the ##i##th rectangle as
##f(x_1^"*") ~~ f(M_1) = sin(x_0+(x_1-x_0)/2) = sin(pi/4) = sqrt2/2,##
##f(x_2^"*") ~~ f(M_2) = sin(x_1+(x_2-x_1)/2) = sin((3pi)/4) = sqrt2/2,##
##f(x_3^"*") ~~ f(M_3) = sin(x_2+(x_3-x_2)/2) = sin((5pi)/4) = -sqrt2/2,##
##f(x_4^"*") ~~ f(M_4) = sin(x_3+(x_4-x_3)/2) = sin((7pi)/4) = -sqrt2/2.##
In the end, what you get from MRAM is the following result:
##color(blue)(int_(0)^(2pi) sin(x)dx ~~ lim_(n->4) sum_(i=1)^4 sin(M_i)Deltax)##
##= (sin(pi/4) + sin((3pi)/4) + sin((5pi)/4) + sin((7pi)/4))*Deltax##
##= (sqrt2/2 + sqrt2/2 - sqrt2/2 - sqrt2/2) * pi/2##
##= color(blue)(0)##
Which is not surprising given that ##sin(x)## in ##[0,pi]## is equal to ##-sin(x)## in ##[pi,2pi]##, which means that
##int_(0)^(pi) sinxdx = -int_(pi)^(2pi) sinxdx##,
and thus:
##color(green)(int_(0)^(2pi) sinxdx)##
##= int_(0)^(pi) sinxdx + int_(pi)^(2pi) sinxdx##
##= color(green)(0)##,
regardless of the chosen method.
Note that RRAM, LRAM, and MRAM are approximations, so it was a coincidence that it gave the exact answer.