(A)

#sum_(k=1)^n8/n log(1+8 (k/n)) = 8sum_(k=1)^nlog(1+8 (k/n)) 1/n#

Wann #n->oo# Wir haben

#lim_(n->oo)sum_(k=1)^nlog(1+8 (k/n)) 1/n approx int_0^1 log(1+8x)dx#

wo für #k=1,2,cdots,n#

#1/n = Deltax#, #0 < (k/n= k Delta x = x_k) le 1 # und

#lim_(Deltax->0)sum_(k=1)^nlog(1+8 x_k) Deltax = int_0^1 log(1+8x)dx#

So

#8int_0^1 log(1+8x)dx = 11.775021196025975#

(B)

Auf die gleiche Weise

#lim_(nrarrinfty)sum_{i=1}^{n}pi/n[sin(pi+(ipi)/n)+2] = pi lim_(nrarrinfty)sum_{i=1}^{n}[sin(pi+pi(i/n))+2]1/n approx#

#approx pi int_0^1 (sin(pi+pi x)+2)dx =2(pi-1) #