Question #0debb

1 Answer
Jan 26, 2018

Let us look at the definition of the Riemann definite integral:
#int_a^bf(x)\ dx=(b-a)lim_(N->oo)1/Nsum_(i=1)^Nf(a+(i(b-a))/N)#

With this, we shall manipulate the sum in question into a form that can be translated to the definite integral.

#\ \ \ \ \ \ lim_(n->oo)sum_(i=1)^N(N+i)/N^2#
#=lim_(n->oo)1/Nsum_(i=1)^N(N+i)/N#
#=lim_(n->oo)1/Nsum_(i=1)^N(1+i/N)#
#=(1-0)lim_(n->oo)1/Nsum_(i=1)^N(1+(0+(i(1-0))/N))#
#=int_0^1(1+x)\ dx#
#=[(1+x)^2/2]_0^1#
#=3/2#