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

1 Answer
Aug 7, 2017

Answer:

# int_(-1)^0 \ x^2-x^3 \ dx = 7/12 #

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)#

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

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

And so:

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

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

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

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

# \ \ = lim_(n rarr oo) 1/n sum_(i=1)^n \ {2-(5i)/n+(4i^2)/n^2-i^3/n^3}#

# \ \ = lim_(n rarr oo) 1/n {2sum_(i=1)^n 1 - 5/nsum_(i=1)^n i + 4/n^2 sum_(i=1)^n i^2-1/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) 1/n {2 n - 5/n 1/2n(n+1) + 4/n^2 1/6n(n+1)(2n+1) - 1/n^3 1/4n^2(n+1)^2 }#

# \ \ =lim_(n rarr oo) 1/n {2n - 5/2(n+1) + 2/(3n)(n+1)(2n+1) - 1/(4n) (n+1)^2 }#

# \ \ =lim_(n rarr oo) 1/(12n^2) {24n^2 - 30n(n+1) + 8(n+1)(2n+1) - 3 (n+1)^2 }#

# \ \ = 1/12 lim_(n rarr oo) 1/(n^2) {24n^2 - 30n^2-30n + 16n^2+24n+8 - 3n^2-6n-3 }#

# \ \ = 1/12 lim_(n rarr oo) 1/(n^2) {7n^2 -6n+5 }#
# \ \ = 1/12 lim_(n rarr oo) {7 -6/n+5/n^2 }#

And this is now a trivial limit to evaluate, so:

# L = 1/12 {7 -0+0 }#
# \ \ = 7/12 #

Using Calculus

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

# int_(-1)^0 \ x^2-x^3 \ dx = [ x^3/3-x^4/4 ]_(-1)^0 #
# " " = (0-0) - (-1/3-1/4) #
# " " = 7/12 #