Question

Setup the limit definition of definite integrals `int_-1^2(1+x^2)dx` ?

Thanks for your help!

Answer 1

`int_(-1)^2 \ (1+x^2) \ dx = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+((3i)/n-1)^2}`

Which if we evaluate gives us `I=6`

Explanation:

Using Riemann sums. 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. The limit of which is known as the Riemann sum (or the limit definition of an integral)

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-1 }`

Here we have `f(x)=1+x^2` and so we partition the interval `[-1,2]` using:

`Delta = {-1, -1+1(3)/n, -1+2 (3)/n, -1+3 (3)/n, ..., 2 }`

And so:

`I = int_(-1)^2 \ (1+x^2) \ dx`

`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(-1+(3i)/n)`

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

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

Which is the desired limit.

We can continue and evaluate the limit (ie the integral) by 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)`

So, we have:

`I = lim_(n rarr oo) 3/n sum_(i=1)^n \ {1+(9i^2)/n^2-(6i)/n+1}`

`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {2+(9i^2)/n^2-(6i)/n}`

`\ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n \ 2 + 9/n^2 sum_(i=1)^n \ i^2 - 6/n sum_(i=1)^n \ i }`

`\ \ = lim_(n rarr oo) 3/n {2n + 9/n^2 1/6n(n+1)(2n+1) - 6/n 1/2n(n+1) }`

`\ \ = lim_(n rarr oo) 3/n {1/(2n)(4n^2 + 3(n+1)(2n+1) - 6n(n+1) }`

`\ \ = lim_(n rarr oo) 3/n {1/(2n)(4n^2 + 3(2n^2+3n+1) - (6n^2+6n) }`

`\ \ = lim_(n rarr oo) 3/(2n^2) {4n^2 + 6n^2+9n+3 - 6n^2-6n }`

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

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

`\ \ = 3/2 \ {4 + 0+0 }`

`\ \ = 6`

Using Calculus

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

`I = int_(-1)^2 \ 1+x^2 \ dx`
`\ \ = [x+x^3/3]_(-1)^2`
`\ \ = (2+8/3)-(-1-1/3)`
`\ \ = 6`