Question

Use Riemann sums to evaluate? : `int_0^(pi/2) \ sinx \ dx` ?

Answer 1

`int_0^(pi/2) \ sinx \ dx = 1`

Explanation:

By definition of an integral, then

`int_a^b \ f(x) \ dx`

represents the area under the curve `y=f(x)` between `x=a` and `x=b`. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

`int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)`

And we partition the interval `[a,b]` equally spaced using:

`Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) }`
`\ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b }`

Here we have `f(x)=sinx` and so we partition the interval `[0,pi/2]` using:

`Delta = {0, 0+1(pi/2)/n, 0+2 (pi/2)/n, 0+3 (pi/2)/n, ..., pi/2 }`

And so:

`I = int_0^(pi/2) \ sinx \ dx`
`\ \ = lim_(n rarr oo) (pi/2)/n sum_(i=1)^n \ f(0+i*(pi/2)/n)`
`\ \ = lim_(n rarr oo) pi/(2n) sum_(i=1)^n \ sin((ipi)/(2n) )` ..... [A}

If we were dealing with polynomials, we would now utilise standard summation formulas for powers, but as we are dealing with a sine summation those results will not help.

In order to evaluate the sine sum, We consider the following:

`1 + z + z^2 + ... + z^n = (z^(n+1)-1)/(z-1) \ \` for z `!= 1`

in combination with Euler's formula by taking

`z = e^(i theta) =cos theta+i sin theta`

Then applying De Moivre's theorem:

`e^(i n theta)=cosntheta+isin n theta`

and equating real and imaginary parts, we eventually find that:

`sum_(j=1)^n sin(jtheta) = (cos(theta/2)-cos(n+1/2)theta) / (2sin(theta/2)`
`" " = (cos(theta/2)-cos(ntheta+theta/2)) / (2sin(theta/2)`
`" " = (cos(theta/2)-(cos ntheta cos(theta/2) - sinnthetasin(theta/2))) / (2sin(theta/2)`

`" " = (cos(theta/2)-cos ntheta cos(theta/2) + sinnthetasin(theta/2)) / (2sin(theta/2))`

So, if we put `theta=(pi)/(2n)` and use this result in the earlier summation [A], we get:

`I = lim_(n rarr oo) pi/(2n) {(cos((pi)/(4n))-cos (pi/2) cos((pi)/(4n)) + sin (pi/2) sin((pi)/(4n))) / (2sin(pi/(4n)))}`

`\ \ = lim_(n rarr oo) pi/(2n) {(cos((pi)/(4n)) + sin((pi)/(4n))) / (2sin(pi/(4n)))}`

`\ \ = pi/4 \ lim_(n rarr oo) 1/(n) (cot((pi)/(4n)) + 1 )`

`\ \ = pi/4 { lim_(n rarr oo) 1/(n) (1/tan((pi)/(4n))) + lim_(n rarr oo) 1/(n) }`

`\ \ = pi/4 { lim_(n rarr oo) 1/(n) (( pi/(4n))/tan((pi)/(4n))) * (4n)/pi + 0 }`

`\ \ = lim_(n rarr oo) ( pi/(4n))/tan((pi)/(4n))`

And a standard calculus limit is:

`lim_(x rarr 0) x/tanx =1`

Utilising this result with `x=pi/(4n)` we get:

`I = 1`

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

`int_0^(pi/2) \ sinx \ dx = [ -cos ]_0^(pi/2)`
`" " = -(cos (pi/2)-cos0)`
`" " = -(0-1)`
`" " = 1`