Use Riemann sums and a limit to compute the exact area under the curve.

**JDS** · December 4th 2012, 08:17 PM

Use Riemann sums and a limit to compute the exact area under the curve.

Y = 4x² – x on (a) [0, 1]; (b) [-1, 1] ; (c) [1,3]

I have already done a and verified it is correct, however b and c I am unsure of. This is as far as I get on b, then I am not sure what to do next....

(b)
∆x = 2/n, We use right endpoints as evaluation points, x_i = -1 + 2i/n

(my small amount of work is shown in the picture I attached.)

Also, I worked this problem out on paper, but the end result seemed to way off, so I started over and only managed to confuse myself.

I guess I am really getting confused in the part where x_i = -1 + 2i/n, where usually it is a single term such as i/n. Anyways any help on this matter is greatly appreciated!

**MarkFL** · December 5th 2012, 01:49 AM

I think what I would do here, is instead of working out 3 separate sums, I would use the interval [a,b], to derive a formula for any interval. Then we can plug in the intervals we are given.

We will have:

$\Delta x=\frac{b-a}{n}$

$x_k=a+k\Delta x$

Hence:

$A_n=\Delta x\sum_{k=0}^{n-1}\left(4x_k^2-x_k \right)=\Delta x\sum_{k=0}^{n-1}\left(4(a+k\Delta x)^2-(a+k\Delta x) \right)=$

$\Delta x\sum_{k=0}^{n-1}\left(4a^2+8a\Delta xk+4(\Delta x)^2k^2-a-\Delta xk \right)=\Delta x\sum_{k=0}^{n-1}\left(a(4a-1)+\Delta x(8a-1)k+4(\Delta x)^2k^2 \right)=$

$\Delta x\left(a(4a-1)n+\Delta x(8a-1)\frac{(n-1)n}{2}+2(\Delta x)^2\frac{(n-1)n(2n-1)}{3} \right)=$

$(b-a)\left(a(4a-1)+\frac{(b-a)(8a-1)(n-1)}{2n}+\frac{2(b-a)^2(n-1)(2n-1)}{3n^2} \right)$

Thus, we find:

$\lim_{n\to\infty}A_n=(b-a)\left(a(4a-1)+\frac{(b-a)(8a-1)}{2}+\frac{4(b-a)^2}{3} \right)=$

$(b-a)\left(\frac{6a(4a-1)+3(b-a)(8a-1)+8(b-a)^2}{6} \right)=$

$(b-a)\left(\frac{8(a^2+ab+b^2)-3(a+b)}{6} \right)=\frac{4(b^3-a^3)}{3}+\frac{a^2-b^2}{2}$

Now it's simply a matter of using the given values to answer parts a), b) and c).

**JDS** · December 5th 2012, 10:27 AM

Thank you MarkFL2!! I just 'tested' your derived formula with my known correct answer for a and it was spot on! You are a life saver! Thank you!!! I will post my answers for b and c in a moment to see if I did them correctly. Thanks again!!

**JDS** · December 5th 2012, 10:45 AM

(b)

(4(1^3 - (-1)^3)/3) + (((-1)^2 - 1^2)/2)
= 8/3 + 0/2.....but doesnt this create an undefined answer or null answer because of 0/2?

or is it simply 8/3?

**JDS** · December 5th 2012, 10:55 AM

(c)

(4(3^3 - 1^3)/3) + ((1^2 - 3^2)/2)

= 104/3 - 8/2

= 92/3

**MarkFL** · December 5th 2012, 12:47 PM

Yes, b) is 8/3 and c) is 92/3.

**JDS** · December 5th 2012, 08:59 PM

Originally Posted by MarkFL2

Thus, we find:

$(b-a)\left(\frac{6a(4a-1)+3(b-a)(8a-1)+8(b-a)^2}{6} \right)=$

$(b-a)\left(\frac{8(a^2+ab+b^2)-3(a+b)}{6} \right)=\frac{4(b^3-a^3)}{3}+\frac{a^2-b^2}{2}$

I am trying to use your method for a different formula, but once I reach the point above, I get stuck.

How are you coming up with that whole last line I quoted. I just don't get how that all factors down to the final equation. I mean it definitely gave me the correct answer for the other problems. But when trying to mimic this formula for a different (but similar equation) I just do not understand how to factor out this last part. Would you be so kind as to explain that last part a bit further? Thank you so much (in advance!) for your taking the time to help me!

**JDS** · December 5th 2012, 09:03 PM

Ah, nevermind....I think I see what mistake I have made....In my calculations I forgot to factor in the (b-a) at the beginning. Makes quite a bit of difference!

**MarkFL** · December 5th 2012, 09:12 PM

Hey, I am glad to see you taking the generalized approach! Not only is it less work in the end, but it forces you to use more algebra up front.

edit: I will be glad to explain any step I made that is not clear.

**MarkFL** · December 5th 2012, 09:55 PM

If I were to look at my set of assigned problems and see several functions of the same type (linear, quadratic, etc.) I might be tempted to generalize not only for the intervals, but for the functions as well.

**JDS** · December 5th 2012, 10:01 PM

Originally Posted by MarkFL2

Hey, I am glad to see you taking the generalized approach! Not only is it less work in the end, but it forces you to use more algebra up front.

edit: I will be glad to explain any step I made that is not clear.

Well, regardless of the "Eureka" moment I had

, I think I may have to take you up on the offer to explain a bit further.

I will post the new equation I am working on and my attempt to work out my derived formula, and the answer that gives.

I will also show my original attempt to solve the problem (before I attempted your better method). As you might already guess, my two answers did not match, which tells me I did one or the other incorrectly!

So here is the equation i have been trying to derive:

y = x² + 3x on [0, 1]

I have attached my work in pic files and they are labeled accordingly.

**MarkFL** · December 5th 2012, 10:18 PM

In your first attempt, I am with you up to this point:

$A_n=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac {3}{n^2}\cdot\frac{n(n+1)}{2}$

Now, I would simplify as follows:

$A_n=\frac{(n+1)(2n+1)}{6n^2}+\frac{3(n+1)}{2n}= \frac{(n+1)((2n+1)+9n)}{6n^2}=\frac{(n+1)(11n+1)}{ 6n^2}$

Now, what is the limit of this sum as $n\to\infty$ ?

I am going to look over your more generalized approach next.

**MarkFL** · December 5th 2012, 10:29 PM

In your second approach, you actually made the same error that you made in your first approach. I am with you up to this point:

$A_n=(b-a)\left(a(a+3)+\frac{(b-a)(2a+3)(n-1)}{2n}+\frac{(b-a)^2(n-1)(2n-1)}{6n^2} \right)$

Now, when you took the limit as $n\to\infty$ you used a denominator of 6 for the last term within the parentheses, when it should be a 3, do you see why?

I will let you rework this on your own once you understand why you need a 3 there. Everything else you did is correct except for that.

edit: If you have a bunch of quadratic functions, you may want to generalize further and use $f(x)=ax^2+bx+c$ on $[d,e]$ .

**JDS** · December 5th 2012, 10:30 PM

Originally Posted by MarkFL2

In your first attempt, I am with you up to this point:

$A_n=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac {3}{n^2}\cdot\frac{n(n+1)}{2}$

Now, I would simplify as follows:

$A_n=\frac{(n+1)(2n+1)}{6n^2}+\frac{3(n+1)}{2n}= \frac{(n+1)((2n+1)+9n)}{6n^2}=\frac{(n+1)(11n+1)}{ 6n^2}$

Now, what is the limit of this sum as $n\to\infty$ ?

I am going to look over your more generalized approach next.

Ah, thank you for the clarification on that part. Simplifying it does help. After doing so, I come up with:

$A_n=\frac{11}{6}\ + \frac{2}{n}\ + \frac{1}{6n^2}$

**JDS** · December 5th 2012, 10:35 PM

Originally Posted by MarkFL2

In your second approach, you actually made the same error that you made in your first approach. I am with you up to this point:

$A_n=(b-a)\left(a(a+3)+\frac{(b-a)(2a+3)(n-1)}{2n}+\frac{(b-a)^2(n-1)(2n-1)}{6n^2} \right)$

Now, when you took the limit as $n\to\infty$ you used a denominator of 6 for the last term within the parentheses, when it should be a 3, do you see why?

I will let you rework this on your own once you understand why you need a 3 there. Everything else you did is correct except for that.

edit: If you have a bunch of quadratic functions, you may want to generalize further and use $f(x)=ax^2+bx+c$ on $[d,e]$ .

I do not see why it should be a 3, at least not yet...I am now reworking the problem at this point and will let you know if I cannot figure out why it should be a 3...thanks for all your help and guidance thus far!

Edit: So, yeah, I see it now....2/6 reduces to 1/3....Im gonna chalk that up to being sleep deprived!

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