A lower level question of curiosity.

[dtl42]

Hello, I am a high school student, and I am a new member to this forum.

I was wondering how to find the area under the graph of one sin arc without using calculus. The problem comes from something one of my friends gave me. The question shows a picture with some rectangles in the space underneath the curve. I was thinking if i drew a large, scaled version and filled it with many rectangles i would be able to approximate the area. the key point to my question is that calculus is not involved.

Are there any methods that are less demeaning than drawing tons of little rectangles? What are they?

 

[JasonRox]

Hello, I am a high school student, and I am a new member to this forum.

I was wondering how to find the area under the graph of one sin arc without using calculus. The problem comes from something one of my friends gave me. The question shows a picture with some rectangles in the space underneath the curve. I was thinking if i drew a large, scaled version and filled it with many rectangles i would be able to approximate the area. the key point to my question is that calculus is not involved.

Are there any methods that are less demeaning than drawing tons of little rectangles? What are they?

Definitely. You can get great approximations without even using calculus.

To get a good approximation you might want to look into Riemann Sums and once you get that look at the Trapezoidal Rule, which is just an improved version. There is also a Simpsons Rule, but I don't quite remember the method. I do believe it was the most accurate of the three though.

In the end, I suggest you go with the Riemann Sums. It's the easiest one to follow, it's very straight forward. In fact, you probably already used Riemann Sums without even knowing it. (It involves rectangles.) Also, what's neat about it is that you can increase the approximation with relative ease compared to an actual drawing.

Then, you will see that the value converges to a number as you approximate it even more and even more. Say, for example the answer is 3, then what I mean by converge is that, let's say the first try you get 2.5, the second 2.7, the third 2.9, then 2.95, then 2.96, and so on.

The answer isn't 3. I decided not to use the answer so that you discover it yourself.

 

[JasonRox]

Oh, and just search on Google for Riemann Sums. Look for a website with nice diagrams and such.

 

[dtl42]

Ok thanks, and for anyone else who wouldn't mind helping, I am in precalculus right now so I don't even know calculus to begin with.

 

[JasonRox]

Ok thanks, and for anyone else who wouldn't mind helping, I am in precalculus right now so I don't even know calculus to begin with.

You don't need to know calculus for this. As long as you know how to draw a graph, and evaluate a function at a point there you're ready to go.

i.e. Evaluate f(x)=sin(x) at Pi. f(Pi) = sin(Pi) = ?

 

[sutupidmath]

http://science.kennesaw.edu/~plaval/applets/Riemann.html [Broken]


just follow this link and u will get a nice applet to help u with it.

 

[dtl42]

Thanks so far everyone, but the only site that explains how to use the Riemann Sums on the calculator is a little confusing, could anyone help with this? I have a Ti- 84 SI+

 

[matt grime]

Since when were Riemann sums not calculus/analysis?

 

[JasonRox]

Since when were Riemann sums not calculus/analysis?

I meant in a way where it doesn't use derivatives and such.

 

[JasonRox]

http://archives.math.utk.edu/visual.calculus/4/riemann_sums.4/

There's a good website to use.

 

[matt grime]

Thanks so far everyone, but the only site that explains how to use the Riemann Sums on the calculator is a little confusing, could anyone help with this? I have a Ti- 84 SI+

It's nothing to do with your calculator. Just draw some rectangles or parallelograms under the curve to estimate the area. Add 'em by hand if you want. You won't get the actual integral, just a numerical estimate of it. The real integral is defined as the limit of the estimates as the width of your rectangles tends to zero, so you cannot, by definition, avoid analysis.

 

[dtl42]

OK thanks you guys, so I've drawn from your responses that the only non-calc/analysis method to do this is with the little rectangles. Looks to be fun time! (10)

 

[JasonRox]

OK thanks you guys, so I've drawn from your responses that the only non-calc/analysis method to do this is with the little rectangles. Looks to be fun time! (10)

That's what Riemann Sums does. It draws little rectangles for you.

Check out the website I linked about. It gives pictures explaining everything.

 

[dtl42]

O yea i know that's what it does but everyone else said that its considered analysis, so i thought i should stay away from it?

 

[dtl42]

Alright I took some time to draw the rectangles and such and i got an answer first of 1.87, then 1.89, then 1.92, and most recently 1.95, would it be safe to assume, following logical order that the area is approaching 2? Is this the correct answer? Thanks very much!

 

[uart]

Alright I took some time to draw the rectangles and such and i got an answer first of 1.87, then 1.89, then 1.92, and most recently 1.95, would it be safe to assume, following logical order that the area is approaching 2? Is this the correct answer? Thanks very much!

No it wouldn't be safe to conclude that based just on the data you have. In general (series approximating) integrals do not necessarily converge to nice round numbers, so assuming "2" for that reason would be fallacy.

In this case however (assuming you are integrating sin(x) from x = zero to Pi) then as it turns out yes the exact answer is 2. It's just that it wouldn't be safe to conclude that from your data alone.

 

[JasonRox]

In general (series approximating) integrals do not necessarily converge to nice round numbers, so assuming "2" for that reason would be fallacy.

Oh really.

Are you sure about that?

 

[matt grime]

yes, Jason. He is. And he is correct. There is nothing posted by dtl42 that implies that that sequence of numbers he wrote converges to 2. It could be 2-sqrt(3)/e^4000.

 

[Gib Z]

yes, Jason. He is. And he is correct. There is nothing posted by dtl42 that implies that that number converges to 2. It could be 2-sqrt(3)/e^4000.

We'll, You could probably tell there's no e in it, seeing as its a sin curve :P Unless someone can prove pi and e are not algebaricly independant...

But I agree with the point, we can't say the data implies a value of 2, after more terms we could say 2 may be a good approximation.

EDIT: "We'll, You could probably tell there's no e in it, seeing as its a sin curve :P Unless someone can prove pi and e are not algebaricly independant..."

If the bounds aren't designed to be fitting to that solution as well lol >.< I doubt this dudes doing that though.

 

[matt grime]

We'll, You could probably tell there's no e in it, seeing as its a sin curve :P Unless someone can prove pi and e are not algebaricly independant...

what has that got to do with anything? If we're going to throw in some off topic comments, then, note that sin(x) = (expix - exp-ix)/2i, so maybe e and i will appear in the answer!

 

[Gib Z]

I just got brandished with a stick >.<
Ill keep my mouth shut next time :)

 

[christianjb]

Weigh a piece of graph paper. Divide by total number of squares to get the weight per square. Take a scissors, cut out the area under the curve. Weigh the cut-out. Divide the weight by the weight per square.

Stand on your head.

 

[dtl42]

Ok So the answer is 2, but i cannot conclude that from my data? And

CHristianjb - I was going to do something kind of like that but with rice? Where i fill it up with rice and then fill up a rectangle with as much rice as i can and then measure the area of that. I believe that's a similar tactic. Thanks a lot everyone for the help!

 

[matt grime]

The answer is 2, but you can *never* conclude that from doing an approximation like this. It might be 1.999...95 and therefore differ from 2 but only at a point far beyond your calculator's capabilities.

 

[JasonRox]

yes, Jason. He is. And he is correct. There is nothing posted by dtl42 that implies that that sequence of numbers he wrote converges to 2. It could be 2-sqrt(3)/e^4000.

I agree with him saying can't it's 2, but many integrals come out with nice numbers. Especially the ones you start out with.

Also, he's implying that if you have data (better data than his) seems to imply that the solution is a nice round number, then something must be wrong. I would never conclude what the solution is using approximations, but you still can't throw out solutions like 2 simply because it's a nice round number.

 

[matt grime]

I agree with him saying can't it's 2, but many integrals come out with nice numbers. Especially the ones you start out with.

and many don't. Far more than do. There are after all uncountably many "non-round numbers", and arguably only countably many "round numbers". It is exactly the same analogy as leads us to say that almost no functions have anti-derivatives.

Also, he's implying that if you have data (better data than his) seems to imply that the solution is a nice round number, then something must be wrong.

No he's not. Sure, 2 is just as likely an answer as any other, in some sense, but that is not what he's saying. If we believe the answers to be uniformly distributed on some interval [a,b], then with probability 1 it will not be a "round number". It is of course open to interpretation what round means in this context. And for the OP to leap from his data to thinking it converges to 2 ignores *uncountably* many other options between his largest approximation and 2, and it is a fallacy to make that leap.

 

[uart]

do not necessarily converge to nice round numbers

Also, he's implying that if you have data (better data than his) seems to imply that the solution is a nice round number, then something must be wrong.

Jason, what part of "not necessarily" don’t you understand? I never said that the solution could not be an integer or simple fraction, I just didn't want OP to jump to the conclusion that this was typically the case based on this one example.

I'm just astounded that a statement as elementary as this, (that definite integrals are not necessarily nice round numbers) could generate any controversy at all. Honestly I would have thought that to be about as controversial as saying that the Easter Bunny doesn't exist!

 

[JasonRox]

Jason, what part of "not necessarily" don’t you understand? I never said that the solution could not be an integer or simple fraction, I just didn't want OP to jump to the conclusion that this was typically the case based on this one example.

I'm just astounded that a statement as elementary as this, (that definite integrals are not necessarily nice round numbers) could generate any controversy at all. Honestly I would have thought that to be about as controversial as saying that the Easter Bunny doesn't exist!

Ok, I agree.

I mentionned that his things should get closer and closer to some value. I gave him that because I knew that. If was one of my earlier posts. The reason why I wanted him to use Riemann Sums was because then he can actually explore what the real value is not just have an approximation. Because then, I would just tell him after he had enough practice, to have the subinterval to 0. As in, evaluate the limit. So, he can explore not only solutions to areas underneath the sin curve but eventually others well, and possible recognize that they don't always converge.