How do you evaluate the definite integral #int x^2 dx# from #[0,1]#?

Question

Elijah Abram · Accepted Answer

I am assuming that you do not yet have the Fundamental Theorem of Calculus available to evaluate this, but that you need to evaluate it from a definition.

.#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_0^1 x^2 dx#. For each #n#, we get #Deltax = (b-a)/n = (1-0)/n = 1/n# And #x_i = a+iDeltax = 0+i1/n = i/n# #f(x_i) = (x_i)^2 = (i/n)^2# # = i^2/n^2# #sum_(i=1)^n f(x_i)Deltax = sum_(i=1)^n(i^2/n^2) 1/n# # = sum_(i=1)^n(i^2/n^3)# # = 1/n^3 sum_(i=1)^n i^2 # # = 1/n^3[(n(n+1)(2n+1))/6]# (We used summation formulas for the sums in the previous step.) So, #sum_(i=1)^n f(x_i)Deltax = 1/n^3[(n(n+1)(2n+1))/6]# # = 1/6[(n(n+1)(2n+1))/n^3]# The last thing to do is evaluate the limit as #nrarroo#. There are a couple of ways to think about this limit : The numerator can be expanded to a plynomial with leading term #2n^3#, so the limit as #nrarroo# is #2#. OR #(n(n+1)(2n+1))/n^3 = (n/n)((n+1)/n)((2n+1)/n)# The limit at infinity is #(1)(1)(2)=2# as a product of rational expressions. To finish the calculuation, we have #int_0^1 x^2 dx = lim_(nrarroo)1/6[(n(n+1)(2n+1))/n^3] = 1/6[2] =1/3.# Using the Fundamental Theorem of Calculus Find an antiderivative of #x^2# (call it #F(x)#) and evaluate #F(1)-F(0)# #int_1^2(x^2) dx = {: (x^3/3)]_0^1# # = 1/3#

Charles Anctil · Answer

undefined To evaluate the definite integral ∫x^2 dx from [0,1], you use the fundamental theorem of calculus. The antiderivative of x^2 is (1/3)x^3. Evaluating this antiderivative at the upper and lower limits of integration (1 and 0, respectively) and subtracting yields the result.

∫x^2 dx from [0,1] = (1/3)x^3 evaluated from 0 to 1.
Plugging in the upper limit:
(1/3)(1)^3 = 1/3
Plugging in the lower limit:
(1/3)(0)^3 = 0

Subtracting the lower limit value from the upper limit value:
1/3 - 0 = 1/3

Therefore, the value of the definite integral ∫x^2 dx from [0,1] is 1/3.