# Reimann sum and summation - Khan Academy

From Khan Academy, we want to approximate the area in the interval $[1,7]$ using a right Riemann sum with $9$ equal subdivisions. Only 1 answer is correct:

My question is, equation $y = \frac5x + 2$ is already given and we can use right Reimann sum and add the areas of different rectangles each of $\frac{7-1}{9} = \frac23$ width. We only need 3 things: $x, \Delta x, y$ and we got all of them, then from where this 2nd equation (the answer with strange interval of $[1,9]$) came up ? Does some 3rd equation exist too with another strange interval $[2,10]$ e.g. ?

Let me be clear:

$$y = \frac 5x + 2, \Delta x = \frac23$$

Hence the $9$ intervals are:

$$[1,\frac53], [\frac53, \frac73], [\frac73, 3], [3, \frac{11}{3}], [\frac{11}{3}, \frac{13}{3}]$$ $$[\frac{13}{3}, 5], [5, \frac{17}{3}], [\frac{17}{3}], \frac{19}{3}], [\frac{19}{3}, 7]$$

Now we can calculate right Reimann sum using right-end values of each interval for $y$ is already given:

$$y(\frac53) = \frac{5}{\frac{5}{3}} + 2 \iff 3 + 2 \iff 5$$

Similarly we can calculate all and add them.

$$y(\frac73) = \frac{29}{7}, y(3) = \frac{11}{3}, y(\frac{11}{3}) = \frac{37}{11}, y(\frac{13}{3}) = \frac{41}{13}$$

$$y(5) = 3, y(\frac{17}{3}) = \frac{49}{17}, y(\frac{19}{3}) = \frac{53}{19}, y(7) = \frac{19}{7}$$

Hence, we got the length and width of all the rectangles and we can dot he summation:

$$Area = \frac23 (5 + \frac{29}{7} + \frac{11}{3} + \frac{37}{11} + \frac{41}{13} + 3 + \frac{49}{17} + \frac{53}{19} + \frac{19}{7})$$

$$Area = 20.475$$

Like I said, we already have $x, \Delta x, y$ and we know how to do summation, then why we need this 2nd equation that shows up as answer ?

The answer is given in the very beginning of the referenced site:

Summation notation can be used to write Riemann sums in a compact way. This is a challenging, yet important step towards a formal definition of the definite integral.

My question is, equation $y=\frac5x+2$ is already given and we can use right Reimann sum and add the areas of different rectangles each of $\frac{7−1}9=\frac23$ width.

Yes, you can, but the objective is given in the next statement on the referenced site:

Summation notation (or sigma notation) allows us to write a long sum in a single expression. While summation notation has many uses throughout math (and specifically calculus), we want to focus on how we can use it to write Riemann sums.

We only need 3 things: $x,Δx,y$ and we got all of them, then from where this 2nd equation (the answer with strange interval of $[1,9]$) came up ? Does some 3rd equation exist too with another strange interval $[2,10]$ e.g. ?

The left and right Riemann sum formulas with intervals are given on the referenced site (in your case $n=9$):

$\hspace{2cm}$

What you did (calculate the functional values on the right borders of the sub-intervals, multiply them by the width of each interval and sum them all) is correct, but you must follow the instruction ("approximate the area in the interval $[1,7]$ using a right Riemann sum with $9$ equal subdivisions").

• Thanks for so much of explanation :) . I missed the "challenging" part on the top (kind of sub-heading). So, it is all just for formal definition. I am more on "applied math" kinda guy, since this is formal-definition, that is the reason I found difficulty understanding it. I just applied $$\sum_{i=1}^n \Delta x \cdot f(x_i)$$ on the values. I still don't understand it, I have never been able to wrap my head about formal definitions and theoretical research in Math but I understand concrete applications well. – Arnuld Sep 11 '18 at 13:50
• it is good that you understand the background of the Riemann sum. Later on the number of subdivisions $n$ will tend to infinity to result in the Riemann integral. So, keep exercising and asking questions. Good luck. – farruhota Sep 11 '18 at 14:09

You seem to be confusing the interval of integration (in this case, $[1,7]$) with the index of summation, $i \in \{1, 2, \ldots, 9\}$. They do not correspond to each other.

The lower and upper indices of summation correspond to two notions here: (1) whether the Riemann sum uses the left endpoints, or the right endpoints; and (2) how many subdivisions are in the Riemann sum.

If the left endpoints are used, the index starts at $i = 0$, and ends at $i = n-1$, where $n$ is the number of subdivisions; in your case, $n = 9$. If the right endpoints are used, then the index starts at $i = 1$, and ends at $i = n$.

For $n$ equal subdivisions of the interval $[a,b]$, the width of each rectangle is, as you pointed out, $$\Delta x = \frac{b-a}{n}.$$ The $x$-values of the endpoints being evaluated are therefore $$x \in \{a + i \Delta x\} = \left\{ a + \frac{b-a}{n} i \right\}$$ where $i$ takes on values as described above. Then $f(x)$ evaluated at these points is simply $$f\left(a + \frac{b-a}{n} i\right).$$ In your case, $a = 1,$ $b = 7$, $n = 9$, and $f = \frac{5}{x} + 2$, thus $$f\left(a + \frac{b-a}{n} i\right) = f\left( 1 + \frac{2}{3} i\right) = \frac{5}{1 + 2i/3} + 2 = \frac{15}{3 + 2i} + 2.$$

• I understand that answer given here works. May be I was not clear and hence I edited the OP to reflect my question. Thing is, why create a formula for a formula ? We have $x$, $\Delta X$ and $y$ and we know how to do summation. Then, we can simply do it, why another formula ? – Arnuld Sep 11 '18 at 6:35