How do you evaluate the definite integral by the limit definition given #int x^3dx# from [-1,1]?

We have,
# int_-1^1x^3dx = [x^4/4]_-1^1 #

# = [(-1)^4/4 - (1)^4/4]#
# = [1/4 - 1/4]#
# = 0#

NB; If you want the AREA bounded by the curve, between x=-1 to x=1 and the x-axis, then the answer is
A=# 2int_0^1x^3dx = 1/2 # (by symmetry)

Here is a limit definition of the definite integral. (I'd guess it's the one you are using.)

.#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_-1^1 x^3 dx#.

Find #Delta x#

For each #n#, we get

#Deltax = (b-a)/n = (1-(-1))/n = 2/n#

Find #x_i#

And #x_i = a+iDeltax = -1+i2/n = -1+(2i)/n#

Find #f(x_i)#

#f(x_i) = (x_i)^3 = (-1+(2i)/n)^3 = -1+6i/n-12i^2/n^2+8i^3/n^3#

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

#sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n( -1+6i/n-12i^2/n^2+8i^3/n^3) 2/n#

# = sum_(i=1)^n( -1/n+6i/n^2-12i^2/n^3+8i^3/n^4)#

# =sum_(i=1)^n ( -1/n)+sum_(i=1)^n(6i/n^2)-sum_(i=1)^n(12i^2/n^3)+sum_(i=1)^n(8i^3/n^4) #

# =-1/nsum_(i=1)^n ( 1)+6/n^2sum_(i=1)^n(i)-12/n^3sum_(i=1)^n(i^2)+8/n^4sum_(i=1)^n(i^3) #

Evaluate the sums

# = -1/n(n) +6/n^2((n(n+1))/2) - 12/n^3((n(n+1)(2n+1))/6) +8/n^4( (n^2(n+1)^2)/4) #

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

Rewrite before finding the limit

#sum_(i=1)^n f(x_i)Deltax = -1/n(n) +6/n^2((n(n+1))/2) - 12/n^3((n(n+1)(2n+1))/6) +8/n^4( (n^2(n+1)^2)/4#

# = -1 +3((n(n+1))/n^2) - 2((n(n+1)(2n+1))/n^3) +2( (n^2(n+1)^2)/n^4)#

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

#lim(nrarroo) ((n(n+1))/n^2) = 1#

#lim(nrarroo) ((n(n+1)(2n+1))/n^3) = 2#

#lim(nrarroo) (n^2(n+1)^2)/n^4) = 1#

To finish the calculation, we have

#int_0^1 x^2 dx = lim_(nrarroo) -1 +3((n(n+1))/n^2) - 2((n(n+1)(2n+1))/n^3) +2( (n^2(n+1)^2)/n^4)#

# = -1+3(1)-2(2)+2(1) = -1+3-4+2 =0#