Integrating by parts:
#int_0^1 xe^xdx = int_0^1 x d(e^x)#
#int_0^1 xe^xdx = [xe^x]_0^1 - int_0^1e^xdx#
#int_0^1 xe^xdx = e - [e^x]_0^1#
#int_0^1 xe^xdx = e - e +1 = 1#
If you want to demonstrate as limit of the Riemann sum, consider the partitions of the interval #[0,1]# in #n# intervals of length #1/n# limited by the points:
#x_(k,n) = k/n# for #k=0,1,...,n#
for every #n in NN#.
The left Riemann sum is then:
#s_n = sum_(k=0)^(n-1) (x_ke^(x^k))/n = sum_(k=1)^(n-1) (ke^(k/n))/n^2 = 1/n^2 sum_(k=1)^(n-1) ke^(k/n) #
While the right Riemann sum is:
#S_n = sum_(k=0)^(n-1) (x_(k+1)e^(x^(k+1)))/n = sum_(k=1)^n (ke^(k/n))/n^2 #
so that:
#(1) " " int_0^1 xe^xdx = lim_(n->oo) 1/n^2 sum_(k=1)^n ke^(k/n) #
Now consider the geometric series:
#1/(1-x) = sum_(k=0)^oo x^k#
converging for #x in (-1,1)# and differentiate it term by term:
#1/(1-x)^2 = sum_(k=1)^oo kx^(k-1) = 1/x sum_(k=1)^oo kx^k#
For #x=e^(1/n)#:
#1/(1-e^(1/n))^2 = 1/e^(1/n) sum_(k=1)^oo k(e^(1/n))^k#
#e^(1/n)/(1-e^(1/n))^2 = sum_(k=1)^oo ke^(k/n)#
Dividing both sides by #n^2# we have:
#e^(1/n) /(n^2(1-e^(1/n))^2) = 1/n^2sum_(k=1)^oo ke^(k/n)#
and comparing this to equation #(1)# we get:
#(2) " " int_0^1 xe^xdx = lim_(n->oo) e^(1/n) /(n^2(1-e^(1/n))^2) #
Using the well known limit:
#lim_(x->0) (e^x-1)/x = 1#
we can see that:
#lim_(n->oo) (e^(1/n)-1)/(1/n) = lim_(n->oo) n(e^(1/n)-1) = 1 #
and then:
#lim_(n->oo) e^(1/n) /(n^2(1-e^(1/n))^2) = lim_(n->oo) e^(1/n) * 1/(lim_(n->oo) n(e^(1/n)-1))^2 = 1#
and we can conclude that:
#int_0^1 xe^xdx = 1#
