Integrating by parts:
int_0^1 xe^xdx = int_0^1 x d(e^x)∫10xexdx=∫10xd(ex)
int_0^1 xe^xdx = [xe^x]_0^1 - int_0^1e^xdx∫10xexdx=[xex]10−∫10exdx
int_0^1 xe^xdx = e - [e^x]_0^1∫10xexdx=e−[ex]10
int_0^1 xe^xdx = e - e +1 = 1∫10xexdx=e−e+1=1
If you want to demonstrate as limit of the Riemann sum, consider the partitions of the interval [0,1][0,1] in nn intervals of length 1/n1n limited by the points:
x_(k,n) = k/nxk,n=kn 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