# Riemann Sum Question

1. Aug 22, 2012

### joe_cool2

*SOLVED*Riemann Sum Question

*SOLVED*

My question is quite simple. I probably just missed something somewhere. I've looked for hours and cannot find the mistake.

1. The problem statement, all variables and given/known data
Find the area under the curve using the definition of an integral and Gauss summation equations:

f(x) = 3 - (1/2)x
where x is greater than or equal to two and less than or equal to 14

2. Relevant equations

Formula #1: Gauss equation for the sum of a list simple list of numbers eg 1,2,3,etc.:

[n(n+1)]/2

Formula #2: to find area using Riemann sums:

lim as n→∞ of Ʃ from i=1 to n of:
f[(i*(b-a))/n]*[(b-a)/n]

3. The attempt at a solution

Using Formula #2:

lim as n→∞ of Ʃ from i=1 to n of:
f[12i/n]*(12/n)

pulling out 12/n from under the summation sign:

lim as n→∞ of 12/n * Ʃ from i=1 to n of:
3 - (6i/n)

pulling 36/n out from underneath the summation sign because it has no "counting" i variable:

lim as n→∞ of 36/n - (72/n^2) * Ʃ from 1 to n of i

Using Formula #1 to get rid of the summation sign:

lim as n→∞ of 36/n - (72n^2 + 72n)/2n^2

crossing out the factors n^2 and 2, which are in the N and D of the 2nd fraction:

lim as n→∞ of 36/n - 36 - 36/n

taking the limit:

-36

Now what's the problem?

142 3 - (1/2)x dx = -12

What went wrong? Again, I've been checking this thing for hours.

Last edited: Aug 22, 2012
2. Aug 22, 2012

### jbunniii

This is evaluating the function in the interval [0,12], not [2,14].

3. Aug 22, 2012

### gustav1139

Yeah you have the formula for the Riemann sum just slightly (but crucially) wrong. It should be:

$\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(a+\frac{i(b-a)}{n})\cdot(\frac{b-a}{n})$

4. Aug 22, 2012

### joe_cool2

Alright. Well here's the updated version using the correct formula.

lim as n→∞ of Ʃ from i=1 to n of:
f[(12i/n)+2]*(12/n)

pulling out 12/n from under the summation sign:

lim as n→∞ of 12/n * Ʃ from i=1 to n of:
3 - (6i/n) - 1

pulling 36/n out from underneath the summation sign because it has no "counting" i variable:

lim as n→∞ of 36/n - [(72/n^2) * Ʃ from 1 to n of i] - 12/n

Using Formula #1 to get rid of the summation sign:

lim as n→∞ of 36/n - (72n^2 + 72n)/2n^2 - 12/n

crossing out the factors n^2 and 2, which are in the N and D of the 2nd fraction:

lim as n→∞ of 36/n - 36 - 36/n - 12/n

taking the limit:

-36

Still the same, wrong answer. Am I just being stupid or what?

Last edited: Aug 22, 2012
5. Aug 22, 2012

### gustav1139

There's two mistakes I see:

when you plugged in f(2+12/n), you flipped a sign.

you can't really "pull out 36/n" the way you have; sorry I didn't catch this first time around.

When you're at the stage where you have $\frac{12}{n}\sum{(3-6i/n)}$, you have to be a bit more careful about how you simplify this sum.

Try breaking it up into two sums for instance.

6. Aug 22, 2012

### Dick

I think gustav1139 already found the problem, but you aren't presenting the solution very clearly either. Go back and change 3 - (6i/n) + 1 to 3 - (6i/n) - 1. That's where you flipped the sign.

7. Aug 22, 2012

### joe_cool2

Okay I changed them to minus signs. I see where I messed up there. But as far as I remember, you can pull factors out of the sums as long as no terms have an "i" in them. Why can I not do it in this case? I basically did split it up like:

(12/n) Ʃ 3 - (12/n) Ʃ (6i/n) - (12/n) Ʃ 1

When I said "pull out 36/n" I meant that the sum from 1 to n of 3 is just 3, so:

36/n - (12/n) Ʃ (6i/n) - (12/n) Ʃ 1

8. Aug 22, 2012

### gustav1139

What's $\sum_{i=1}^{n}3$?

9. Aug 22, 2012

### joe_cool2

Oh, dear. It's 3n, isn't it? Sorry, it's been about six months since the beginning of my Calc II class last semester. Thanks for your patience.

EDIT: Solved.

Last edited: Aug 22, 2012