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

Question

Emma Aldridge · Accepted Answer

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

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 #

Christopher Agresta · Answer

undefined To evaluate the integral $ \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx $ using Riemann sums, you can use the definition of the Riemann sum:

$$ \sum_{i=1}^{n} f(x_i) \Delta x $$

where $ \Delta x $ is the width of each subinterval and $ x_i $ are the sample points within each subinterval. For this integral, you can choose $ n $ subintervals of equal width.

Choose $ n $ subintervals of equal width $ \Delta x = \frac{\pi}{2n} $. Then, the sample points for each subinterval will be $ x_i = \frac{\pi i}{2n} $ for $ i = 1, 2, ..., n $.

Now, evaluate $ f(x_i) = \sin(x_i) $ at each sample point and multiply by $ \Delta x $. Then, sum up all these products from $ i = 1 $ to $ n $.

$$ \sum_{i=1}^{n} \sin\left(\frac{\pi i}{2n}\right) \frac{\pi}{2n} $$

As $ n $ approaches infinity, this Riemann sum approaches the value of the integral. So, to evaluate the integral, take the limit as $ n $ approaches infinity of this sum.