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

1 Answer
Apr 18, 2017

Answer:

# I = int_-1^1 \ (1-x^3) \ dx =2 #

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)=1-x^3# and we partition the interval #[-1,1]# using #Delta = {-1, -1+2*1/n, -1+2*2/n, ..., -1+2*n/n }#

And so:

# I = int_-1^1 \ (1-x^3) \ dx #
# \ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ f(-1+2*i/n)#
# \ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ (1-(-1+(2i)/n)^3) #
# \ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ (1-(-1 + 3((2i)/n) - 3((2i)/n)^2+((2i)/n)^3)) #

# \ \ = lim_(n rarr oo) 2/n sum_(i=1)^n \ (2 - (6i)/n + (12i^2)/n^2-(8i^3)/n^3) #

# \ \ = lim_(n rarr oo) 2/n {sum_(i=1)^n2 - 6/nsum_(i=1)^n i + 12/n^2sum_(i=1)^ni^2-8/n^3 sum_(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) 2/n {2n - 6/n 1/2n(n+1) + 12/n^2 1/6n(n+1)(2n+1)-8/n^3 1/4n^2(n+1)^2}#

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

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

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

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

# \ \ = lim_(n rarr oo) 2/n^2 {-n^2-3n + 4n^2+6n+2-2n^2-4n-2}#

# \ \ = lim_(n rarr oo) 2/n^2 {n^2-n}#

# \ \ = lim_(n rarr oo) (2-2/n)#
# \ \ = 2#