0
$\begingroup$

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:

enter image description here enter image description here

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 ?

$\endgroup$
0
$\begingroup$

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.

Your said:

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.

Your said:

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 $n=9$):

$\hspace{2cm}$enter image description here

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").

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Arnuld Sep 11 '18 at 13:50
  • $\begingroup$ 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. $\endgroup$ – farruhota Sep 11 '18 at 14:09
1
$\begingroup$

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 $$f\left(a + \frac{b-a}{n} i\right) = f\left( 1 + \frac{2}{3} i\right) = \frac{5}{1 + 2i/3} + 2 = \frac{15}{3 + 2i} + 2.$$

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Arnuld Sep 11 '18 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.