Question

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]?

Answer 1

`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`