How do you evaluate the definite integral by the limit definition given #int xdx# from [0,4]?

Here is a limit definition of the definite integral. (I hope it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks.

.#int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#.

Where, for each positive integer #n#, we let #Deltax = (b-a)/n#

And for #i=1,2,3, . . . ,n#, we let #x_i = a+iDeltax#. (These #x_i# are the right endpoints of the subintervals.)

I prefer to do this type of problem one small step at a time.

#int_0^4 x dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (4-0)/n = 4/n#

Find #x_i#

And #x_i = a+iDeltax = 0+i4/n = (4i)/n#

Find #f(x_i)#

#f(x_i) = x_i = (4i)/n#

Find and simplify #sum_(i=1)^n f(x_i)Deltax # in order to evaluate the sums.

#sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ( (4i)/n) 4/n#

# = sum_(i=1)^n( (16i)/n^2)#

# = 16/n^2 sum_(i=1)^n(i)#

Evaluate the sums

# = 16/n^2((n(n+1))/2)#

(We used a summation formula for the sums in the previous step.)

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = 16/n^2((n(n+1))/2)#

# = 8((n(n+1))/n^2))#

Now we need to evaluate the limit as #nrarroo#.

#lim_(nrarroo) ((n(n+1))/n^2) = lim_(nrarroo) (n/n*(n+1)/n) = 1#

To finish the calculation, we have

#int_0^4 x dx = lim_(nrarroo) ( 8((n(n+1))/n^2))#

# = 8(1) = 8#