Hey all, I'm back!

[Lizzie]

Ok, more Calc problems, but these seem quite a bit more managable ones. I think I even know the answers to most of them!! Ok, here's my first problem.

The problem -
Sketch the region enclosed by y=x+1 and y=x². Decide whether to integrate with respect to x or to y. Draw a typical approximating rectangle and label it's height and width. Then find the area of the region.

My steps -
I first sketched the two graphs and the region they enclose.
I dont know if I am being asked which integration works best. I think I have to figure that out before I draw the approximating rectangle.

The next question is very similar and if I can figure out what to do for this one, then I will have no problem with that one. Any help would be very much appreciated. Thanks!

 

[stapel]

Yes, you're being asked to decide which way, x-axis-based or y-axis-based, looks easier to integrate. (There isn't necessarily a "right" answer to this, though there is likely an "expected" answer. But as long as you get the right integral value, you should be fine.)

It's your choice. Find the intersection points, (a, b) and (c, d). Then decide if you want to integrate between y = x + 1 and y = x² (from a to c) or between x = sqrt[y] and x = y - 1 (from b to d) plus between x = sqrt[y] and x = 0 (from 0 to b).

Once you decide which direction to take, start working on your Riemann sum.

Eliz.

 

[galactus]

Hello Lizzie, welcome back.

An interesting thing about this particular problem is the intersection points of the two functions.

Solve \(\displaystyle x^{2}=x+1\) for x. The two solutions are interesting in that they are what is known as the "Golden Ratio". You could use these values as your limits of integration.

 

[Lizzie]

I feel stupid asking this, but what exactly does that mean galactus? And thank you both for your help.

 

[stapel]

what exactly does that mean galactus?

Which part?

The Golden Ratio is just a curiosity that he noticed when he was setting up how to find the intersection points of y = x² and y = x + 1, assuming you're going with an x-axis-based orientation.

It's just FYI; it's not required for the integration.

Eliz.

 

[galactus]

I just mentioned it for curiosity's sake. It isn't directly pertinent to your problem but just an observation. The "Golden Ratio" has many implications. If you want to learn more, do a google search. It's a rather interesting topic and too vast for me to get into on this thread.

 

[Lizzie]

lol, ok, I will check that out. So, what kind of steps would I take to do this? I may have missed it if you wrote it already, so I will check again.
Thanks again!

 

[stapel]

So, what kind of steps would I take to do this?

1) Pick an orientation: x-axis-based (traditional) or y-axis-based (flipped on its side).

2) Write the equations in terms of that orientation: "y=" for x-axis-based and "x=" for y-axis-based.

3) Find the intersection points of the two equations.

4) Set up the integral. The specifics will depend on the orientation you choose, but the general set-up will be:

. . . . .\(\displaystyle \large{\int^b_a{\mbox{ (top line's eqn.) }-\mbox{ (bottom line's eqn.) }}dx}\)

...where "a" is the lower limit (left-hand interval endpoint) and "b" is the upper limit (right-hand interval endpoint).

Eliz.

 

[Lizzie]

Alrighty, that helped alot. I will get started right away. I guess I would be better off not waiting until the last moment to do this stuff, but lately it just seems like I keep getting so much school work shoved at me plus everythign at home...Argh! I'm sure you've all been through it at one point in time or another, lol.

 

[stapel]

It's always a rush between Thanksgiving and Xmas. Sometimes it's like every instructor thinks that he's the only one who thought to give you an over-the-holiday assignment. Ugh!

:wink: :lol:

Eliz.

 

[Lizzie]

The bad part is, I didn't get a break for Thanksgiving because I take online courses. So, i still had work to do over the holiday.

 

[Lizzie]

1) Pick an orientation: x-axis-based (traditional) or y-axis-based (flipped on its side). I chose x-axis-based.

2) Write the equations in terms of that orientation: "y=" for x-axis-based and "x=" for y-axis-based. Would it be the same as I was given? Like this: y=x +1 and y=x²?

3) Find the intersection points of the two equations. Would these be (-.6,.4) and (2,3)?

4) Set up the integral. The specifics will depend on the orientation you choose, but the general set-up will be:

. . . . .\(\displaystyle \large{\int^b_a{\mbox{ (top line's eqn.) }-\mbox{ (bottom line's eqn.) }}dx}\)

ok, let me see if I can do this since I have your code right here. . .
. . . . .\(\displaystyle \large{\int^3_._4{\mbox{ (I'm not sure which equation you mean by top line's equation) }-\mbox{ (I also don't know what you mean by bottom line's equation) }}dx}\)

...where "a" is the lower limit (left-hand interval endpoint) and "b" is the upper limit (right-hand interval endpoint).

Eliz.

Where do I got from here, if I even have any of that right...

 

[stapel]

1) That's the orientation I would choose. It saves you having to break the integral into to parts.

2) Since you will be working with y in terms of x, yes, stick with the "y in terms of x" equations.

3) As mentioned earlier, the solutions are the Golden Ratio. While that particular piece of information isn't necessary to the exercise, the fact that the Golden Ratio involves irrational numbers containing square roots should have hinted at the fact that these lines do not intersect at whole-number values.

How did you get "x = -3/5" and "x = 2" for your solutions to the Quadratic Formula?

4) Look at the drawing. Which equation draws the top line? Which equation draws the bottom line?

Eliz.

 

[Lizzie]

wow, i was such a dummy because after I looked over what I wrote, I realized a few things that I had done wrong. I'll go back and try to work out what I can.

 

[Lizzie]

Alright, so, I was sitting at the kitchen table last night actually reading and trying my hardest to comprehend this stuff...and I actually understand a large bit of it now! I just have to work on my integrating skills because honestly, I've been cheating with that. I've been using this site that does it for me...I'm a bad girl. Now, I have to figure out how to do it by myself. So, I figure if I get that down I can do this as well.

 

[Lizzie]

As mentioned earlier, the solutions are the Golden Ratio. While that particular piece of information isn't necessary to the exercise, the fact that the Golden Ratio involves irrational numbers containing square roots should have hinted at the fact that these lines do not intersect at whole-number values.
Ok, with a bit of help, I realized what I had to do. Now, I have for my a and b a= (1-sqrt(5))/2, b=(1+sqrt(5))/2
Look at the drawing. Which equation draws the top line? Which equation draws the bottom line? The top line is drawn by y=x+1 and the bottom is drawn by y=x²

Eliz.

So, now I have all this information. I know my a, my b, and my eqauation. So, now I need to intergrate [-x²+x+1], right? That is: x+ x²/2-x³/3

And then I just plug in a and b?
That would be: a=-.1573786517, b=1.515028324
What do I do from here? You said to get started on my reimann sum?

 

[stapel]

What do I do from here? You said to get started on my reimann sum?

Assuming you're still working with Riemann sums.... Have you moved on to using the integration rules directly, instead?

Eliz.

 

[Lizzie]

well, the problem said to "draw a typical approximating rectangle and label it's width and height."

btw, I just read the LaTex thing because I realized that I had to switch my style to the new one instead of SwiftBlue. So, I can show you what my actual updated problem looks like. or at least try.

\large{\int^{1+sqrt{5}/2}_{1-sqrt{5}/2}{}[-x^2+x+1]{}dx}

and then I just used that handy dandy integrator thing and got

\large{x+{{x^2}/2}-{{x^3}/3}

Into which I put a and b and got what I said before.

*How come that didn't work, since you can see exactly the text that I wrote...where did I go wrong in the coding?

 

[stapel]

well, the problem said to "draw a typical approximating rectangle and label it's width and height."

Yes, but you can draw a rectangle. I was referring to the last bit, which asks you to find the area. For this, you need either the Riemann sum or the integration rules.

*How come that didn't work, since you can see exactly the text that I wrote...where did I go wrong in the coding?

It doesn't look like you enclosed anything in the "\(\displaystyle " and "\)" tags. If you don't tell the script that you're posting LaTeX, it doesn't process it that way.

Eliz.

 

[Lizzie]

thank you eliz