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 ?
$\endgroup$2 Answers
$\begingroup$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 $$f\left(a + \frac{b-a}{n} i\right) = f\left( 1 + \frac{2}{3} i\right) = \frac{5}{1 + 2i/3} + 2 = \frac{15}{3 + 2i} + 2.$$
$\endgroup$ 1 $\begingroup$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 $n=9$):
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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").
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