Solve Riemann Sum Question for Area Under Curve

[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.

Homework Statement

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

Homework 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]

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.

 

[jbunniii]

Using Formula #2:

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

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

 

[gustav1139]

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

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

 

[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?

 

[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.

 

[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.

 

[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

 

[gustav1139]

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

 

[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.