# [Integrals] Just can't quite finish this

1. Aug 31, 2004

### Caldus

Hi,

I am trying to solve this problem:
Find the right rieman sum and the integral (using the definition of integral) of:
(-x^2/3)-7 on the interval [0,4] using n subintervals.

So:
lim n->infinity of [sigma from i = 1 to n of (-(4i/n)^2/3)-7]*4/n

= lim n->infinity of [sigma from i = 1 to n of (-4i^2/3n^2)-7]*4/n
= lim n->infinity of [sigma from i = 1 to n of i^2 - 7]*(-16/4n^3)
= lim n->infinity of [sigma from i = 1 to n of i^2] - 7n*(-16/4n^3)
= lim n->infinity of ((n(n + 1)(2n + 1))/6)-7n*(-16/4n^3)

I don't how to go any farther with this...

Thanks for any help.

2. Aug 31, 2004

### Hurkyl

Staff Emeritus
Well, you have lots of techniques for working with limits. For instance, you can usually break lim (A + B) into lim A + lim B. You can combine your fractions into a single fraction, and use what you know about limits of rational functions. If you think the result looks unreasonable, you can check your algebra. Et cetera.

3. Aug 31, 2004

### Caldus

OK here we go:

= lim n->infinity of ((n(n + 1)(2n + 1))/6)-7n*(-16/4n^3)
= lim n->infinity of ((2n^3+3n^2+n)/6)+(112n/4n^3)
= lim n->infinity of ((2n^3+3n^2+673n)/24n^3)
= (1/12)+(1/8n)+(673/24n^2)
= 1/12

I know this is wrong though...

4. Sep 1, 2004

### Galileo

I had a hard time decyphering what you typed. It's a bit difficult to oversee.
Try using LaTeX. (Click on the equations to see how it is used.)
It's much more neat too.

Anyway, you have:
$$\sum_{i=1}^{n}f(x_i)\Delta x$$
$$f(x)=-\frac{x^2}{3}-7$$
$$x_i=\frac{4i}{n}$$
$$\Delta x=\frac{4}{n}$$

So the sum is:
$$\sum_{i=1}^{n}\left(-\frac{16i^2}{3n^2}-7\right)\frac{4}{n}=-\frac{64}{3n^3}\sum_{i=1}^{n}i^2-\frac{28}{n}\sum_{i=1}^{n}1$$
Here I've just split the sum into two in the last step. (I think you got this right).

Now for the first sum, use the identity:
$$\sum_{i=1}^{n}i^2=\frac{1}{6}n(n+1)(2n+1)$$
So that:
$$-\frac{64}{3n^3}\sum_{i=1}^{n}i^2=-\frac{64}{3n^3}\frac{1}{6}n(n+1)(2n+1)=-\frac{64}{18n^2}(2n^2+3n+1)$$
Then take the limit $n\rightarrow \infty$
The first sum becomes $-\frac{64}{9}$

The second sum is easy.

Simple integration of the first term gives:
$$\int_0^4\frac{-x^2}{3}dx = \frac{-x^3}{9}|^4_0=-\frac{64}{9}$$
Which matches the sum.

PS: I`m glad we got the old smilies back, the new ones looked silly.

Edit: Corrected a silly mistake

Last edited: Sep 1, 2004
5. Sep 1, 2004

### Hurkyl

Staff Emeritus
Your calculus seems correct: you just need to work on your algebra skills. I think it would pay off for you to dig out your old texts and review.

The big thing you're missing here is the distributive law: (a + b) c = a c + b c, which holds for sums also $c \sum (a + b) = c \sum a + c \sum b$.

In particular:

lim n->infinity of [sigma from i = 1 to n of (-4i^2/3n^2)-7]*4/n
= lim n->infinity of [sigma from i = 1 to n of i^2 - 7]*(-16/4n^3)

When you factor something out of a sum, you have to factor it out of every term, not just one of them.

lim n->infinity of [sigma from i = 1 to n of i^2 - 7]*(-16/4n^3)
= lim n->infinity of [sigma from i = 1 to n of i^2] - 7n*(-16/4n^3)

Conversely, when you expand, each term gets multiplied by the outside constant. Also, you can't just pull additive constants out of a sum. Remember that $\sum (a + b) = \sum a + \sum b$

lim n->infinity of ((2n^3+3n^2+n)/6)+(112n/4n^3)
= lim n->infinity of ((2n^3+3n^2+673n)/24n^3)

And don't forget that when you add fractions that you have to set up a common denominator before you can simply add the numerators. 1/2 + 3/5 is not 4/10... it's 5/10 + 6/10 = 11/10

6. Sep 1, 2004

### Caldus

Thanks guys I eventually figured out the rest of it!

I have another simple algebra problem (doing another problem like this now). How do I isolate the variable i here?:

lim n->infinity 5/n*sigma i = 1 to n of [(25-(5i/n)^2)^(1/2)]

Thanks.

7. Sep 1, 2004

### Hurkyl

Staff Emeritus
You don't.

To do this limit, I think you either need to do some estimation (and prove that the error goes to zero as n goes to infinity), or find some other way to evaluate the integral.

8. Sep 2, 2004

### Caldus

Ah I noticed it finally. The area under that curve is a quarter circle with radius 5! Wow I'm so dumb sometimes. I got it now. Thanks. :)