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I have been given this rather simple looking assignment, which is confusing me a lot.

Given is: $f:[0,5] \rightarrow \mathbb{R}$ where $f(x)=2x+3$.

The first thing I had to do, was to determine the area under the curve, for the given interval $[0,5]$, from the left and right side, for 30 subintervals. I solved this by using $$ S_J = \sum_{j = 1}^{J} f(\xi) \Delta x_j $$

What I now have to solve is: Find the exact values for the riemann-summ in the interval $[0,5]$ with $n$ subintervals, from the left and right side

My initial thought was to do this, to find the exact area:

$$ S_J = \int_{a}^{b} f(x) \, dx $$ But that how would that allow me to do it from the left and right side?

Any ideas? I am completely lost

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    $\begingroup$ I think you're missing the point of the assignment. You shouldn't be using the integral at all. The point of the assignment is to sum an arithmetic progression to find the general form of the left and right Riemann sums in terms of $n$. You will then observe that as $n \to \infty$, both of these sums will converge to the integral. $\endgroup$ Commented Mar 21 at 17:31
  • $\begingroup$ The "left and right side" probably refer to the tagged partitions where you take either the left or right endpoints of each interval as the tag. $\endgroup$
    – Adayah
    Commented Mar 21 at 17:34
  • $\begingroup$ You need to find formulas for $S_N = \Delta \sum_{j = 1}^{N} f(x_j)$ when $x_1=a$ and when $x_1=a+\Delta$ $\endgroup$
    – Vasili
    Commented Mar 21 at 17:36

1 Answer 1

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If you divide the interval $[0,5]$ into $n$ equal subintervals, they have endpoints at $x_j = \frac{5j}{n}$, for each $j$ from $0$ to $n$.

So you have to evaluate the sums $$\begin{aligned} L_J &= \sum_{j=1}^J f(x_{j-1}) \frac{5}{n} \\ U_J &= \sum_{j=1}^J f(x_j) \frac{5}{n} \end{aligned}$$ The idea is to find a closed-form expression for the sum in terms of $J$ alone. If this kind of series is unfamiliar to you, you might consider reading about sums of arithmetic progressions (from Wikipedia) or researching the formula: $$ \sum_{j=1}^J j = \frac{J(J+1)}{2} $$

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  • $\begingroup$ Thanks, I get how these sums are created, but im not sure what to do from here. I dont see any obvious next steps $\endgroup$
    – N G
    Commented Mar 21 at 17:59
  • $\begingroup$ @NG I added a possible next step $\endgroup$ Commented Mar 21 at 18:20

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