Question

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]?

Answer 1

`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`