How do you use the limit process to find the area of the region between the graph #y=x^2+1# and the x-axis over the interval [0,3]?

1 Answer
Sep 7, 2017

Answer:

# int_0^3 \ x^2+1 \ dx = 12 #

Explanation:

By definition of an integral, then

# int_a^b \ f(x) \ dx #

represents the area under the curve #y=f(x)# between #x=a# and #x=b#. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

# int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)#

And we partition the interval #[a,b]# equally spaced using:

# Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) } #
# \ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b } #

Here we have #f(x)=x^2+1# and so we partition the interval #[0,3]# using:

# Delta = {0, 3/n, 2. 3/n,3. 3/n, ..., 3 } #

And so:

# I = int_0^3 \ x^2+1 \ dx #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(0+i*3/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f((3i)/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {((3i)/n)^2+1} #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ { (9i^2)/n^2 + 1 } #
# \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n \ (9i^2)/n^2 + sum_(i=1)^n \ 1 } #
# \ \ = lim_(n rarr oo) 3/n {9/n^2sum_(i=1)^n \ i^2 + sum_(i=1)^n \ 1 } #

Using the standard summation formula:

# sum_(r=1)^n a \ = an #
# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #

we have:

# I = lim_(n rarr oo) 3/n {9/n^2 1/6n(n+1)(2n+1)+n } #

# \ \ = lim_(n rarr oo) 3 {3/(2n^2) (2n^2+3n+1)+1 } #

# \ \ = lim_(n rarr oo) 3/(2n^2) {3 (2n^2+3n+1)+2n^2 } #

# \ \ = lim_(n rarr oo) 3/(2n^2) {6n^2+9n+3+2n^2 } #

# \ \ = lim_(n rarr oo) 3/(2n^2) {8n^2+9n+3 } #

# \ \ = lim_(n rarr oo) (24n^2+27n+9 ) /(2n^2) #

# \ \ = lim_(n rarr oo) (12+27/(2n) + 9/(2n^2)) #

# \ \ = 12+0 + 0 #

# \ \ = 12 #

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

# int_0^3 \ x^2+1 \ dx = [ x^3/3 + x ]_0^3 #
# " " = (9+3)-0 #
# " " = 12#