Question

Use Riemann sums to evaluate? : `int_0^3 \ x^2-3x+2 \ dx`

Answer 1

`int_0^3 \ x^2-3x+2 \ dx = 3/2`

Explanation:

We are asked to evaluate:

`I = int_0^3 \ x^2-3x+2 \ dx`

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.

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^2-3x+2` and we partition the interval `[0,3]` using:

`Delta = {0, 1*3/n, 2*3/n, ..., n*3/n }`

And so:

`I = int_0^3 \ (x^2-3x+2) \ dx`
`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(0+i*(3-0)/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)^2-3((3i)/n)+2}`
`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ {(9i^2)/n^2-(9i)/n+2}`

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

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

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)`

we have:

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

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

`\ \ = lim_(n rarr oo) 3/n 1/(2n) {3 (n+1)(2n+1) - 9n (n+1) + 4n^2}`

`\ \ = lim_(n rarr oo) 3/(2n^2) {3 (2n^2+3n+1) - 9(n^2+1) + 4n^2}`

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

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

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

`\ \ = 3/2 {1+0 -0}`

`\ \ = 3/2`

Using Calculus

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

`I = int_0^3 \ x^2-3x+2 \ dx`
`\ \ = [x^3/3-(3x^2)/2+2x]_0^3`
`\ \ = (9-27/2+6) - (0)`
`\ \ = 3/2`