Question

Find the riemann sum for `f(x)=x+x^2`?

Answer 1

`int_a^b f(x) "d"x = lim_(n->oo) 1/n^2sum_(i=an)^(bn) (i+i^2/n)`

Explanation:

A Riemann sum is a certain type of approximations for an integral. The easiest and simplest method to use is a rectangle sum; where we sum the area of `n` rectangles with equal width to approximate the area under a curve; as `n` approaches infinity, the sum approaches the true value of the net area.

Here's a Riemann sum with `4` rectangles only:

![https://www.khanacademy.org/math/old-ap-calculus-ab/ab-accumulation-riemann-sums/modal/a/left-and-right-riemann-sums]()

This is just an example. Instead, consider `n` rectangles.

Let `f` be a function. Say, we wish to find the area under `f` from `a` to `b`. For the moment, let's only do the case `a=0` and `b=1`. We will generalise later.

Now; imagine `x_0, x_1, x_2, ... x_n` to be some "marks" on the x-axis such that the width of the i-th rectangle is `x_i - x_(i-1)` and `x_0 = 0`.

Let `Delta_i = x_i-x_(i-1)` be the width of the i-th rectangle. As we want `Delta_i` to be constant, we can more easily call it `color(red)(Delta)`.

Another property about our x-axis marks is that the lenght/non-width side of the i-th rectangle is going to be `f(x_i)`. Hence, the its area is going to be

`"Area"_i = "width" xx "lenght" = Deltaf(x_i)`

Since we want the area until `b=1`, we must have `x_n=1`.

The areas of the rectangles approximate the areas under the graph of `f`. If there are `n` rectangles, then:

`int_0^1 f(x) "d"x ~~ sum_(i=1)^n (x_i-x_(i-1))f(x_i) = sum_(i=1)^n Deltaf(x_i)`

One way to make `Delta_i` constant is by defining `x_i = i/n`. Then,

`Delta_i = i/n - (i-1)/n = 1/n`, `forall i in NN`.

By this definition, `x_0 = 0` and `x_n = 1`, as desired.

`int_0^1f(x) "d"x ~~ sum_(i=1)^n 1/n f(i/n)`

The more rectangles we have, the better the sum approximates the integral. Thus, we can say that

`int_0^1 f(x) "d"x = lim_(n->oo) sum_(i=0)^n 1/n f(i/n)`

Generalisation: Since we want the area from `a` to `b`, the very first rectangle must have lenght `f(a)` and the last one `f(b)`; to do this, we can change the lower and upper bounds of the sum operator:

`color(blue)(int_a^b f(x) "d"x = lim_(n->oo) sum_(i=an)^(bn) 1/n f(i/n))`

This is one of the many Riemann Sum formulae for a function `f`.

In our case, `f(x) = x+x^2`:

`:. int_a^b f(x) "d"x ~~ sum_(i=an)^(bn) 1/n (i/n + i^2/n^2)`

`:. int_a^b f(x) "d"x ~~ 1/n^2sum_(i=an)^(bn)(i+i^2/n)`

if there are `n` rectangles, as stated already.