What is the integral of sin(x) dx from 0 to 2pi?

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An choice way to do this starting from the name determination is:

##int_(a)^(b) f(x)dx = lim_(n->oo) sum_(i=1)^N f(x_i^"*")Deltax##

where:

  • ##n## is the enumerate of rectangles used to abut the total, i.e. the area between the deflexion and the x-axis.
  • ##i## is the refutation of each rectangle in ##[0,2pi]##.
  • ##N## is the refutation of the definite rectangle in ##[0,2pi]##.
  • ##f(x_i^"*")## is the acme of each abandoned rectangle in ##[0,2pi]##, which varies as ##sin(x)##.
  • ##Deltax## is the width of each abandoned rectangle in ##[0,2pi]##, which converges to ##0## as ##n->oo##.

If we use the midpoint-rectangular similarity course (MRAM), we elect a commodious interim ##Deltax## such that we can confront a midpoint for each rectangle of mass ##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 elect ##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 acme 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 forthcoming 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##

##= tint(blue)(0)##

Which is not portentous abandoned that ##sin(x)## in ##[0,pi]## is resembling to ##-sin(x)## in ##[pi,2pi]##, which instrument 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##

##= tint(green)(0)##,

regardless of the chosen course.

Note that RRAM, LRAM, and MRAM are similaritys, so it was a correspondence that it gave the just reply.


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