Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 29

You have to recall the definition of the Reiman Integral int_a^bf(x)dx=lim_(n->oo)sum_(i=1)^nf(x(i))Deltax
where Deltax =(b-a)/n and x(i)= a +iDeltax
x
a=2 and b = 6
Deltax = (6-2)/n= 4/n
x(i) = 2 + i4/n
f(x(i))= (2+i(4/n))/(1+(i(4/n))^5)
Simplifying it will be
f(x(i)) = 2/(1+i^5(4/n)^5)+i(4/n)/(1+i^5(4/n)^5)
f(x(i))=2/(1+(i^5(1024))/n^5)+((4i)/n)/(1+(i^5(1024))/n^5)

f(x(i))=(2n^5)/(n^5+1024i^5)+((4i)(n^4))/(n^5+1024i^5)

Thus,
lim_(n->oo)sum_(i=1)^n((2n^5)/(n^5+1024i^5)+((4i)(n^4))/(n^5+1024i^5))(4/n)