# Thread: Remain Sum (What am I doing for this)

1. ## Remain Sum (What am I doing for this)

I haven't got to the lesson in class about this, but I'd like to jump start before I learn anything in class about it.

The Question:

Consider the Integral

$\displaystyle \int_{2}^{6}(\frac{4}{x} +5)$

(a) Find the Riemann sum for this integral using right endpoints and n=4.
(b) Find the Riemann sum for this same integral, using left endpoints and n=4
To do this, am I taking the anti-derivative to get f(x) then solving for part a and b?

The Anti-derivative by the way is:
$\displaystyle 4ln(\left|x \right|)+5x$

Lastly, this has nothing to do with the problem, but the thread its self. What tags should I put on this to help someone else in the future?

2. Originally Posted by Zanderist
I haven't got to the lesson in class about this, but I'd like to jump start before I learn anything in class about it.

The Question:

Consider the Integral

$\displaystyle \int_{2}^{6}(\frac{4}{x} +5)$

(a) Find the Riemann sum for this integral using right endpoints and n=4.
(b) Find the Riemann sum for this same integral, using left endpoints and n=4

To do this, am I taking the anti-derivative to get f(x) then solving for part a and b?
No, I don't think that this is what you are asked to do.

The Riemann sum for this integral using right endpoints would be something like

$\displaystyle \sum\limits_{k=1}^n f\left(a+k\cdot\frac{b-a}{n}\right)\cdot \frac{b-a}{n} =\sum\limits_{k=1}^n f\left(2+k\cdot\frac{6-2}{n}\right)\cdot \frac{6-2}{n}$

and using left endpoints it would be

$\displaystyle \sum\limits_{k=1}^n f\left(a+(k-1)\cdot\frac{b-a}{n}\right)\cdot \frac{b-a}{n}=\sum\limits_{k=1}^n f\left(2+(k-1)\cdot\frac{6-2}{n}\right)\cdot \frac{6-2}{n}$

where $\displaystyle f(x) := \frac{4}{x}+5$.

Just search your textbook, your lecture notes, or Wikipedia for "Riemann sum", if the above formulas seem completely foreign to you.

Now plug in the given value of n.