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

Question

Christopher Allers · Accepted Answer

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

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:

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.

Ezekiel Alexander · Answer

undefined To find the Riemann sum for $ f(x) = x + x^2 $, you need to first specify the interval over which you want to compute the Riemann sum and the method of partitioning that interval. Let's assume we want to find the Riemann sum for $ f(x) = x + x^2 $ over the interval $[a, b]$, and we'll use $ n $ subintervals of equal width.

The width of each subinterval will be $ \Delta x = \frac{b - a}{n} $. Then, the endpoints of the subintervals are $ x_0 = a, x_1, x_2, ..., x_{n-1}, x_n = b $, where $ x_i = a + i\Delta x $ for $ i = 0, 1, ..., n $.

Now, for each subinterval $ [x_{i-1}, x_i] $, choose a sample point $ c_i $ within that interval. The Riemann sum is then given by:

$$ \sum_{i=1}^{n} f(c_i) \Delta x $$

where $ f(c_i) $ is the value of the function $ f(x) $ evaluated at the sample point $ c_i $ within the $ i $th subinterval.

Therefore, the Riemann sum for $ f(x) = x + x^2 $ over the interval $[a, b]$ with $ n $ subintervals of equal width is:

$$ \sum_{i=1}^{n} (c_i + c_i^2) \Delta x $$

where $ \Delta x = \frac{b - a}{n} $, and $ c_i $ is any point in the $ i $th subinterval.