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

Question

Luke Alvarez · Accepted Answer

#int_0^1 xe^xdx = 1#

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#

Elijah Abram · Answer

undefined To solve the integral $ \int_{0}^{1} x e^{x} \, dx $ using the limit of a sum (which is essentially the definition of the definite integral), you can approximate the integral using Riemann sums. We'll use right Riemann sums in this case.

1. Divide the interval $[0, 1]$ into $n$ subintervals of equal width $ \Delta x = \frac{1}{n} $.

2. Choose sample points $ x_i $ in each subinterval. We'll choose $ x_i = i \Delta x $ for the right Riemann sum.

3. The right Riemann sum is then given by:
   $$ \sum_{i=1}^{n} f(x_i) \Delta x $$
   where $ f(x) = x e^x $.

4. Substitute $ x_i = i \Delta x $ and $ \Delta x = \frac{1}{n} $:
   $$ \sum_{i=1}^{n} (i \Delta x) e^{i \Delta x} \frac{1}{n} $$

5. Simplify the expression inside the sum and the sum itself:
   $$ \frac{1}{n} \sum_{i=1}^{n} i e^{i \Delta x} $$

6. Now, take the limit as $ n $ approaches infinity:
   $$ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} i e^{i \Delta x} $$

This limit represents the value of the integral $ \int_{0}^{1} x e^{x} \, dx $. Calculating this limit exactly is challenging, so typically you would use numerical methods to approximate it.