How do you find the area between the given curve #y= x^2# and the x-axis given in the interval [0,1]?

1 Answer
Nov 8, 2015

Answer:

The area is #1/3#

Explanation:

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 x^2 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^2dx#.

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_1)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]#

So,

#sum_(i=1)^n f(x_1)Deltax = 1/6 [(n(n+1)(2n+1))/n^3]#

The last thing to do is evaluate the limit as #nrarroo#.
I hope it is clear that this amounts to evaluating
#lim_(nrarroo)(n(n+1)(2n+1))/n^3#

There are several ways to think about this:

Limit of a Rational Expression

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.

OR

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

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

So the limit is, again #2#.

However we get it, we get

.#int_0^1 x^2 dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax#

# = lim_(nrarroo) sum_(i=1)^n((i^2)/n^2) 1/n#

# = lim_(nrarroo) [1/6[(n(n+1)(2n+1))/n^3]]#

# = 1/6 (2)#

# = 1/3#