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

1 Answer
Sep 16, 2016

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.

Explanation:

.#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#