1. ## calculus summations

i missed class today because of a field trip, so i'm completely lost. if anyone could help explain how to do these that would be greatly appreciated! thank you. I don't need an answer, i just need to know how to set up summations

1.) use the Left Endpoint Rectangle Approximation Method to approximate the area
integral from 0 to 2 of x^3 dx
and the Right Endpoint Rectangle Approximation Method for the same problem

2.) integral from 1 to 4 of (x)^(1/2) dx using the trapezoidal rule, left endpt RAM, right endpt RAM, midpt RAM

3.) integral from 0 to pi of cosx dx
trapezoidal rule, left endpt RAM, right endpt RAM, and midpt RAM

2. Maybe this will help, someone will explain it better soon no doubt

When doing the LRAM summation you choose a width of x for your rectangles (like 1/2) and then you cut up the graph into sections. The find the area of each rectangle by multiplying base times height. With the Left RAM you choose the y value at the top left corner of each rectangle. So the sum is

width*height (height is the y value at the left top corner)
width is always whatever you chose at the beginning like 1/2 for example

1/2 * 0 + 1/2 * (1/2)^2 + 1/2 * (1)^2 + 1/2 * (3/2)^2

Hopefully that helps a little, I am not sure exactly what you meant

3. There are formulae for setting these up:

left rectangle rule:$\displaystyle \frac{b-a}{n}\sum_{i=0}^{n-1}f(a+i\frac{b-a}{n})$

right rectangle rule$\displaystyle \frac{b-a}{n}\sum_{i=1}^{n}f(a+i\frac{b-a}{n})$

trapezoidal rule
$\displaystyle \frac{b-a}{n}\sum_{i=1}^{n-1}f(a+i\frac{b-a}{n})+\frac12 f(a)+\frac12 f(b)$

midpoint rule
$\displaystyle \frac{b-a}{n}\sum_{i=0}^{n-1}f(a+\frac{b-a}{2n}+i\frac{b-a}{n})$

I suggest you look at the google image search for midpoint rule, trapezoid rule, left rectangle approx and right rectangle approx to get a better idea of what these are doing.

4. Probably meant for this thread:
thank you! im still confused on summations though.
f(x)= 1/(x+1) from [0,1] and n=2
how do i set up the summation for LRAM, RRAM, MRAM, and the trapezoidal method?
EDIT: Sorry all of those formulas are wrong. I will fix them now
Edit: fixed now. No wonder you were having trouble
Sorry, I forgot to say what all of the letters stood for.
(a,b) is the x-interval you are finding the area above.
n is the number of subintervals you divide that interval into.

So for the left rectangle rule you should get
$\displaystyle \frac{b-a}{n}\sum_{i=0}^{n-1}f(a+i\frac{b-a}{n})$
$\displaystyle =\frac{1-0}{2}\sum_{i=0}^{2-1}f(0+i\frac{1-0}{2})$
$\displaystyle =\frac{1}{2}\sum_{i=0}^{1}f(i\frac{1}{2})$

The other rules are similar

5. thank you i actually got the 1/2 part so i think i'm getting the hang of it
but inside the parenthesis i had (1)/[(x/2)+1] is that the same thing as what you had?

6. thank you i actually got the 1/2 part so i think i'm getting the hang of it
but inside the parenthesis i had (1)/[(x/2)+1] is that the same thing as what you had?
Yes, assuming you used x instead of my i like this:

$\displaystyle \sum_{x=0}^1\frac{1}{\frac{x}{2}+1}$

7. yes thats what i got, i think i understand it better now thank you for your help