How do you use the limit process to find the area of the region between the graph #y=64-x^3# and the x-axis over the interval [1,4]?

1 Answer
May 25, 2017

Answer:

# int_1^4 \ 64-x^3 \ dx = 513/4#

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)#

Here we have #f(x)=64-x^3# and we partition the interval #[1,4]# using:

# Delta = {1, 1+1*3/n, 1+2*3/n, ..., 1+n*3/n } #

And so:

# I = int_1^4 \ (64-x^3) \ dx #
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+i*3/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+(3i)/n)#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {64 - (1+(3i)/n)^3}#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {64 - (1+3((3i)/n)+3((3i)/n)^2+((3i)/n)^3)}#
# \ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {63-(9i)/n-(27i^2)/n^2-(27i^3)/n^3}#

# \ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n 63 - 9/n sum_(i=1)^ni-27/n^2 sum_(i=1)^n i^2-27/n^3sum_(i=1)^n i^3 }#

Using the standard summation formula:

# sum_(r=1)^n r \ = 1/2n(n+1) #
# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #
# sum_(r=1)^n r^3 = 1/4n^2(n+1)^2 #

we have:

# I = lim_(n rarr oo) 3/n {63n - 9/n 1/2n(n+1)-27/n^2 1/6n(n+1)(2n+1)-27/n^3 1/4n^2(n+1)^2 }#

# \ \ = lim_(n rarr oo) 3/n {63n - 9/2 (n+1)-9/(2n)(2n^2+3n+1)-27/(4n)(n^2+2n+1) }#

# \ \ = lim_(n rarr oo) 3/n 1/(4n) { 252n^2 - 18 n(n+1)-18(2n^2+3n+1)-27(n^2+2n+1) }#

# \ \ = lim_(n rarr oo) 3/(4n^2) { 252n^2 - 18n^2-18n-36n^2-54n-18-27n^2-54n-27 }#

# \ \ = lim_(n rarr oo) 3/(4n^2) { 171n^2 -123n-27 }#

# \ \ = 3/4 \ lim_(n rarr oo) { 171 -123/n-27/n^2 }#

# \ \ = 3/4 \ (171 -0-0) #

# \ \ = 513/4 #

Using Calculus

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

# int_1^4 \ 64-x^3 \ dx = [ 64x-1/4x^4 ]_1^4 #
# " " = (256-64)-(64-1/4) #
# " " = 192 -255/4#
# " " = 513/4#