# Calc II

1. Sep 3, 2006

### tony873004

Welcome to my first of many questions.

Express the area under the curve y=x5 from 0 to 2 as a limit.

My guess based on class notes:
$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 0}^2 {\left( {x_i^* } \right)} ^5 \Delta x$$

Back of the book:
$$\mathop {\lim }\limits_{n \to \infty } \frac{{64}}{{n^6 }}\sum\limits_{i = 1}^n {i^5 }$$

What am I missing??
where did $$\frac{{64}}{{n^6 }}$$ come from?

2. Sep 3, 2006

### HallsofIvy

Staff Emeritus
Your expression has $\lim_{n\rightarrow \infty}$ but there is no n in the expression itself.

3. Sep 3, 2006

### tony873004

$$\mathop {\lim }\limits_{\Delta x \to 0} \sum\limits_{i = 0}^2 {\left( {x_i^* } \right)} ^5 \Delta x$$

4. Sep 3, 2006

### 0rthodontist

You want to sum up rectangles on the interval (0, 2) that approximate the function. Your book has chosen to divide up the interval into equal-spaced rectangles. The locations of the edges of their rectangles are
0, 2/n, 4/n, 6/n, ... 2n/n
(Do you see how this evenly divides up the interval (0, 2)?)
For each rectangle, they want to multiply its width by its height to get its area.
-What are the widths of these rectangles?

The height of each rectangle is the value of the function at some point on the interval of the base of the rectangle. In this case they chose the rightmost points of these intervals, i.e.
2/n, 4/n, ... 2n/n
-How can you express the i'th number in this sequence?
-What is the value of the function--the heights of the rectangles--at the i'th point in the sequence?
-What is the area of the i'th rectangle?

Then they want to sum up the areas of the n rectangles.
-What is that sum?

Then they take the limit as n goes to infinity, so that the rectangles get finer and finer and more closely approximate the function's area.
-How do they simplify it to their final answer?

Last edited: Sep 3, 2006
5. Sep 3, 2006

### tony873004

The widths of the rectangles would be (2-0) / n = 2/n

This is where I get confused. If I express it as "i" and I break it into 1000 intervals, then I'd be including 1000^5 in my sumation, which would clearly be wrong since 2^5 should be the largest thing I include in my sumation. I'd have to express that it as "x". Since (b-a) is 2, each "x" would represent i*2/n
so this should be (i*2/n)^5 * (2/n)

$$\sum\limits_{i = 1}^n {\left( {\frac{{2i}}{n}} \right)^5 \frac{2}{n}}$$

$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{2i}}{n}} \right)^5 \frac{2}{n}}$$

not a clue

6. Sep 4, 2006

### 0rthodontist

yep

The thing is that you might have 1000^5 in the numerator, but if you break (0, 2) into 1000 intervals, your denominator--which depends on the number of intervals, n--will also be large.

Right. Actually, isolate the part of this that corresponds to the height of the rectangle and try plugging in some values for n and i--you should see for yourself that it won't exceed 2^5 = 32 so long as i <= n.

Well, you have everything right up until now. What can you do inside the summation just using algebra? Then since n doesn't depend on i, if you have a factor that only involves n you can bring it outside the summation.

Last edited: Sep 4, 2006
7. Sep 4, 2006

### tony873004

$$\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {\frac{{2i}}{n}} \right)^5 \frac{2}{n}} \\ \\ \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{2^5 i^5 }}{{n^5 }} \cdot \frac{2}{n}} \\ \\ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{32i^5 }}{{n^5 }} \cdot \frac{2}{n}} \\ \\ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\frac{{64i^5 }}{{n^6 }}} \\ \\ = \mathop {\lim }\limits_{n \to \infty } \frac{{64}}{{n^6 }}\sum\limits_{i = 1}^n {i^5 } \\ \end{array}$$
Thanks for stepping me through that. Intuitively I get the concept. It only took me about 3 minutes to get the actual answer for the area by writing a short computer program:
Code (Text):
Private Sub Command1_Click()
a = 0 'starting x value
b = 2 'ending x value
n = Val(Text2.Text) 'experiment with different numbers of intervals
t = 0
deltaN = (b - a) / n 'width of each interval

For k = a To b Step deltaN 'sum it up
t = t + deltaN * k ^ 5 ' width * height of each interval
Next k

Text1.Text = t 'output the answer
End Sub

Part (b) of this question is "use a computer algebra system to find the sum in your expression from part (a)." I would have guessed that this meant to come up with an answer, such as 10.667, like I got from the above program provided n was large. But the back of the book gives $$\frac{{n\left( {n + 1} \right)^2 \left( {2n^2 + 2n - 1} \right)}}{{12}}$$
So I guess when they say "find the sum" they mean find an equation, not find the answer?

For the equation, I would have guessed
$$\frac{{64}}{{n^6 }}\left( {1^5 + 2^5 + 3^5 + ...n^5 } \right) \]$$.

I don't see anything in the book that shows how to do it their way. Is that because it
s too complicated to do by hand, hence their instruction to use the computer? Or is this something I should have remembered from algebra class? What should I look up to review this kind of problem?

Finally, part (c) of this question is "Evaluate the limit in part (a)". The back of the book gives 32/3, which agrees nicely with the 10.6667 I got using the compute program. But how would I do it their way to get the answer in fraction form?

Most people in our class took Calc I at SFSU using the same textbook. I took Calc I at the community college. We didn't cover sumations or integration or differentiation of trig functions, so everyone else is two chapters ahead of me. This question wasn't even an assigned homework question. It was just a question I picked at random from the previous chapter in an effort to catch up to the rest of the class.

so... stay tuned for more q's :)

Last edited: Sep 4, 2006