How to solve this using limit of sum? int_0^1xe^xdx10xexdx

1 Answer
Jun 27, 2018

int_0^1 xe^xdx = 110xexdx=1

Explanation:

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^xdx10xexdx=[xex]1010exdx

int_0^1 xe^xdx = e - [e^x]_0^110xexdx=e[ex]10

int_0^1 xe^xdx = e - e +1 = 110xexdx=ee+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