# Evaluating integrals

• Feb 4th 2009, 10:04 AM
fattydq
Evaluating integrals
I've attached a picture of the problem I'm working on involving evaluating integrals, I know how to do it visually (if I have a drawn graph in front of me or can draw a graph) which is how I got the correct answer to #1 that you see there, but when the numbers get too huge too simply graph I have no idea how to go about solving these kinds of problems, I've looked through my notes and book and can't make sense of all the steps they're giving me, could someone help me out?
• Feb 4th 2009, 10:32 AM
fattydq
I apologize for the double post but I figured it'd be better to post this related tidbit here rather than start a whole new thread

What about evaluating integrals from pictures of graphs like this other picture I've attached below, I thought I had it down because the area of a circle is just pi r^2 and I did that and took half of it since it's a half circle, why doesn't that work here, and what should I do?
• Feb 4th 2009, 10:45 AM
running-gag
Quote:

Originally Posted by fattydq
I've attached a picture of the problem I'm working on involving evaluating integrals, I know how to do it visually (if I have a drawn graph in front of me or can draw a graph) which is how I got the correct answer to #1 that you see there, but when the numbers get too huge too simply graph I have no idea how to go about solving these kinds of problems, I've looked through my notes and book and can't make sense of all the steps they're giving me, could someone help me out?

Hi

For the second exercise $\displaystyle \Delta x = \frac{6}{n}$ and $\displaystyle x_i = 1 + \frac{6}{n}\:i$

$\displaystyle \sum_{i=1}^{n}f(x_i) \Delta x = \frac{6}{n} \sum_{i=1}^{n}\left( \left(1 + \frac{6}{n}\:i\right)^2 + 2 \left(1 + \frac{6}{n}\:i\right) - 7 \right)$

$\displaystyle \sum_{i=1}^{n}f(x_i) \Delta x = 120 + \frac{180}{n} + \frac{36}{n^2}$

The integral is the limit when n tends towards $\displaystyle +\infty$ therefore 120
• Feb 4th 2009, 10:53 AM
Jester
Quote:

Originally Posted by fattydq
I've attached a picture of the problem I'm working on involving evaluating integrals, I know how to do it visually (if I have a drawn graph in front of me or can draw a graph) which is how I got the correct answer to #1 that you see there, but when the numbers get too huge too simply graph I have no idea how to go about solving these kinds of problems, I've looked through my notes and book and can't make sense of all the steps they're giving me, could someone help me out?

$\displaystyle \Delta x = \frac{7-1}{n} = \frac{6}{n}$ and $\displaystyle x_i = 1 + \frac{6}{n} i$

so the sum is

$\displaystyle \lim_{n \to \infty} \sum_{i=1}^n \left[ \left( 1 + \frac{6}{n} i \right)^2 + 2 \left(1 + \frac{6}{n} i \right) - 7 \right] \frac{6}{n}$

after expanding

$\displaystyle 6 \lim_{n \to \infty} \sum_{i=1}^n \frac{36 i^2}{n^3} + \frac{24i}{n^2} - \frac{4}{n}$

distributing the sum

$\displaystyle 6 \lim_{n \to \infty} \left( \frac{36}{n^3} \sum_{i=1}^n i^2 + \frac{24}{n^2} \sum_{i=1}^n i - \frac{4}{n} \sum_{i=1}^n 1 \right)$

substituting in the formulas for the sums, expanding and taking the limit gives

$\displaystyle 6 \lim_{n \to \infty} \left( \frac{36}{n^3} \frac{n(n+1)(2n+1)}{6} + \frac{24}{n^2} \frac{n(n+1)}{2} - \frac{4}{n} n \right) = 120$
• Feb 4th 2009, 12:18 PM
fattydq
Quote:

Originally Posted by running-gag
Hi

For the second exercise $\displaystyle \Delta x = \frac{6}{n}$ and $\displaystyle x_i = 1 + \frac{6}{n}\:i$

$\displaystyle \sum_{i=1}^{n}f(x_i) \Delta x = \frac{6}{n} \sum_{i=1}^{n}\left( \left(1 + \frac{6}{n}\:i\right)^2 + 2 \left(1 + \frac{6}{n}\:i\right) - 7 \right)$

$\displaystyle \sum_{i=1}^{n}f(x_i) \Delta x = 120 + \frac{180}{n} + \frac{36}{n^2}$

The integral is the limit when n tends towards $\displaystyle +\infty$ therefore 120

I don't understand at all how either of you got from setting the problem up to the next step where it suddenly equals 120+whatever+whatever. In other words, including "Hi" as a line, I don't understand how you got from line 3 of your work to line 4
• Feb 4th 2009, 01:04 PM
Krizalid
He just set up the sum and after some algebraic expansion he applied properties of sums and that's all. You should know that.
• Feb 4th 2009, 01:06 PM
fattydq
Quote:

Originally Posted by Krizalid
He just set up the sum and after some algebraic expansion he applied properties of sums and that's all. You should know that.

I must not, because it makes no sense how he went from (1+6i/n)^2 to 120, etc etc
• Feb 4th 2009, 01:55 PM
fattydq
Wait, does it have something to do with the n(n+1)/2 formula for summation? It seems like it's connected somehow I just don't know how
• Feb 4th 2009, 02:40 PM
Jester
Quote:

Originally Posted by danny arrigo

$\displaystyle \Delta x = \frac{7-1}{n} = \frac{6}{n}$ and $\displaystyle x_i = 1 + \frac{6}{n} i$

so the sum is

$\displaystyle \lim_{n \to \infty} \sum_{i=1}^n \left[ \left( 1 + \frac{6}{n} i \right)^2 + 2 \left(1 + \frac{6}{n} i \right) - 7 \right] \frac{6}{n}$ (1)

after expanding

$\displaystyle 6 \lim_{n \to \infty} \sum_{i=1}^n \frac{36 i^2}{n^3} + \frac{24i}{n^2} - \frac{4}{n}$ (2)

distributing the sum

$\displaystyle 6 \lim_{n \to \infty} \left( \frac{36}{n^3} \sum_{i=1}^n i^2 + \frac{24}{n^2} \sum_{i=1}^n i - \frac{4}{n} \sum_{i=1}^n 1 \right)$ (3)

substituting in the formulas for the sums, expanding and taking the limit gives

$\displaystyle 6 \lim_{n \to \infty} \left( \frac{36}{n^3} \frac{n(n+1)(2n+1)}{6} + \frac{24}{n^2} \frac{n(n+1)}{2} - \frac{4}{n} n \right) = 120$ (4)

In the four step in red, which one confuses you?
• Feb 4th 2009, 03:20 PM
fattydq
Quote:

Originally Posted by danny arrigo
In the four step in red, which one confuses you?

Well in step 2 for starters you pull the 6 out in front of everything but not the whole 6/n, I don't understand how/why that is. Also in step 2 I don't know how you got the 24i over n^2 and -4/n, I know the 36i^2 comes from 6i being squared but don't understand how you got to the other two terms.

And since there's all that confusion that early on, the rest of your steps subsequently make no sense to me.