Question

Write the Riemann sum to find the area under the graph of the function `f(x) = x^2` from `x = 1` to `x = 5`?

Answer 1

`int_1^5 \ x^2 \ dx = 124/3`

Explanation:

We are asked to evaluate:

`I = int_1^5 \ x^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)`

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)=x^2` and so we partition the interval `[1,5]` using:

`Delta = {1, 1+1(4)/n, 1+2 (4)/n, 1+3 (4)/n, ..., 4 }`

And so:

`I = int_1^5 \ (x^2) \ dx`
`\ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ f(1+(4i)/n)`
`\ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ (1+(4i)/n)^2`
`\ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ ((n+4i)/n)^2`
`\ \ = lim_(n rarr oo) 4/n sum_(i=1)^n \ 1/n^2(n+4i)^2`
`\ \ = lim_(n rarr oo) 4/n^3 sum_(i=1)^n \ (n^2+8ni+16i^2)`
`\ \ = lim_(n rarr oo) 4/n^3 {n^2 sum_(i=1)^n 1 + 8n sum_(i=1)^ni+16sum_(i=1)^ni^2)`

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) 4/n^3 {n^3 + 8n 1/2n(n+1)+16 1/6n(n+1)(2n+1)`

`\ \ = lim_(n rarr oo) 4/n^3 \ n/3 {3n^2 + 12n(n+1)+ 8(n+1)(2n+1)`

`\ \ = lim_(n rarr oo) 4/(3n^2) (3n^2 + 12n^2+12n+ 16n^2+24n+8)`

`\ \ = lim_(n rarr oo) 4/(3n^2) (31n^2+36n+ 8)`
`\ \ = 4/3 lim_(n rarr oo) (31+36/n+ 8/n^2)`
`\ \ = 4/3 (31+0+ 0)`
`\ \ = 124/3`

Using Calculus

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

`I = int_1^5 \ x^2 \ dx`
`\ \ = [x^3/3]_1^5`
`\ \ = 125/3-1/3`
`\ \ = 124/3`