0
$\begingroup$

In a particular question in Calculus, I am asked to find $\int_{-1}^1 x^3$ using a Riemann Sum. The difficulty I am having is picking viable values for $c_i$.For example: I know the value $\frac in -1$ does not work as it values is never greater than 0 for all $i$. However, it is also true that $x_{i-1}\le c_i\le x_i$. Does $c_i$ have to be such that when $i=n$, $c_i=x_n=b$ and when $i=0$, $c_i=x_0=a$?

EDIT: Thank you, all of you who have taken the time to reply. However, I am not having any trouble whatsoever finding the actual integral (I know the fundamental theorem of Calculus and how to find antiderivatives. My question is solely on clarification of choosing the value of $c_i$ in a Riemann Sum

  • 0
    Seriously ($ = 0$) ? Or you meant $\int_0^1 x^3 dx$ – 2017-02-08
  • 0
    Yes. I meant what I wrote above. And no, the question itself is not complicated. I just want to make sure I develop a good understanding of the theory and application or Riemann sums. – 2017-02-08
  • 0
    if $f(x) = -f(-x)$ is continuous then $\int_{-a}^a f(x)dx = \lim_{N \to \infty} \displaystyle\sum_{n=-N}^N \frac{f(na /N)}{a/N} = 0$ – 2017-02-08
  • 0
    I have edited my question to make it more clear what I am asking... hopefully. – 2017-02-08
  • 0
    You didn't write any Riemann sum, I did. I can write also $\displaystyle\int_{-a}^a f(x)dx = \lim_{c \to \infty} \sum_{n, \ n/c \in [- a,a]} \frac{f(n/c)}{c} = 0$ – 2017-02-08
  • 0
    I don't understand your statement about $\frac in-1$ not working. But no, there is no special requirement on the first and last $c_i$. Like all the others, they just have to live in their allotted subinterval. In this case, I would recommend taking advantage of the symmetry and place the $c_i$ symmetrically about the origin, making each Riemann sum equal to zero. Then taking the limit is rather boring. – 2017-02-08
  • 0
    The problem is that the limit of when $c_i = \frac in - 1$ the limit evaluates to $\frac 12$. You can also prove it doesn't work by setting n = to some value and noting that for any given value of $i$ (when $i\le n$) $\frac in -1 \not\ge 0$ – 2017-02-08
  • 0
    @user1952009 I apologize. I didn't include a Riemann sum because a) The Riemann sum was not given in the question b) I assumed that whoever answered the question would be familiar with Riemann Sums, making inclusion of one unnecessary. – 2017-02-08
  • 0
    Ah, but the integral runs from $-1$ to $+1$, so surely, $i$ should range from $0$ to $2n$? (Give or take one at either end.) Or better, just use $c_i=i/n$ with $i$ ranging from $-n$ to $n$. There is nothing magical about having $1$ or $0$ as the lower limit in a sum. By taking $i$ from $0$ to $n$, you are in effect computing the integral from $-1$ to $0$. – 2017-02-08

0 Answers 0