# I can't think of any other way on this

#### allegansveritatem

##### Full Member
Here is the problem:

I tried to solve this by adding the areas of those parts of the figure that seemed to contain whole squares:

The sign before the 12 is supposed to be the sign for "approximates". I suppose in calculus there are ways of working with this kind of problem but I have no calculus at all so...

#### Dr.Peterson

##### Elite Member
Here is the problem:
View attachment 11029
I tried to solve this by adding the areas of those parts of the figure that seemed to contain whole squares:

View attachment 11030

The sign before the 12 is supposed to be the sign for "approximates". I suppose in calculus there are ways of working with this kind of problem but I have no calculus at all so...
Yes, this can be done exactly with calculus (and is probably intended to lead you in the direction of some calculus concepts); but they only want an estimate, and they tell you that you can do different things to get different levels of accuracy.

What you did is a very good first try; but where does the 4 come from?

One way you can improve this is to find one set of squares that is entirely contained within the figure, and another that contains the figure. (That is, one is below and the other is above.) Then you can be sure the the actual area is between the two numbers.

Another thing you could do is to find a figure made of straight lines that is as close as you can come to the actual figure. You might draw lines between the points on the curve, and then add up rectangles and triangles, or trapezoids, to find the area, which will be a closer overestimate.

See what you can do; as they say, it's meant to be a "critical thinking activity", and an opportunity to "be creative", so just have fun with it!

#### allegansveritatem

##### Full Member
Yes, this can be done exactly with calculus (and is probably intended to lead you in the direction of some calculus concepts); but they only want an estimate, and they tell you that you can do different things to get different levels of accuracy.

What you did is a very good first try; but where does the 4 come from?

One way you can improve this is to find one set of squares that is entirely contained within the figure, and another that contains the figure. (That is, one is below and the other is above.) Then you can be sure the the actual area is between the two numbers.

Another thing you could do is to find a figure made of straight lines that is as close as you can come to the actual figure. You might draw lines between the points on the curve, and then add up rectangles and triangles, or trapezoids, to find the area, which will be a closer overestimate.

See what you can do; as they say, it's meant to be a "critical thinking activity", and an opportunity to "be creative", so just have fun with it!
I think your second suggestion is a good way to go: The figure can be broken up into a rectangle with the area 2 x 3. and the upper part of the figure can be made into a right triangle with the area (3 x 9)/2. With this approach I get 19.5 which seems too big to me, at least measured against my first estimate. I wonder if the sag in the hypotenuse could account for some of the contrast here?

#### Dr.Peterson

##### Elite Member
I think your second suggestion is a good way to go: The figure can be broken up into a rectangle with the area 2 x 3. and the upper part of the figure can be made into a right triangle with the area (3 x 9)/2. With this approach I get 19.5 which seems too big to me, at least measured against my first estimate. I wonder if the sag in the hypotenuse could account for some of the contrast here?
That's another estimate that's a little rougher than the first, but still reasonable.

Try tightening it up by using all the points you're shown on the curve, not just the top and bottom ...

#### Jomo

##### Elite Member
Here is the problem:
View attachment 11029
I tried to solve this by adding the areas of those parts of the figure that seemed to contain whole squares:

View attachment 11030

The sign before the 12 is supposed to be the sign for "approximates". I suppose in calculus there are ways of working with this kind of problem but I have no calculus at all so...
No!, the sign before the 12 should be an EQUAL sign simply because 2*3+1*2+4*1 is EXACTLY 12. Here is what you meant to say: Let A be the exact area. Then A ~ 2*3 + 1*2 +4*1 = 12. Do you see the difference?

Last edited:

#### allegansveritatem

##### Full Member
That's another estimate that's a little rougher than the first, but still reasonable.

Try tightening it up by using all the points you're shown on the curve, not just the top and bottom ...
Yes, I see what you mean. In that case I would use one rectangle, two trapezoids, and one right triangle I will work this out later tonight and post. It is not that hard but I am in the middle of something right now. But I see a little more light at the end of the tunnel here.

#### allegansveritatem

##### Full Member
No!, the sign before the 12 should be an EQUAL sign simply because 2*3+1*2+4*1 is EXACTLY 12. Here is what you meant to say: Let A be the exact area. Then A ~ 2*3 + 1*2 +4*1 = 12. Do you see the difference?
Right. I see it. Thanks

#### allegansveritatem

##### Full Member
I divided this figure into a right triangle, two trapezoids and a rectangle and got this:

I guess you might call this procedure an example of "poor man's calculus". But, I doubt if I could get any more accurate than this without further instruction.

#### Dr.Peterson

##### Elite Member
I divided this figure into a right triangle, two trapezoids and a rectangle and got this:
View attachment 11056

I guess you might call this procedure an example of "poor man's calculus". But, I doubt if I could get any more accurate than this without further instruction.
That's as good as you can do using only the points shown on the graph. In calculus, it's equivalent to what is called the "trapezoidal approximation". (It can be done with only three trapezoids, each with altitude 1.) Good work.

The actual area is exactly 15, as it turns out.

#### allegansveritatem

##### Full Member
That's as good as you can do using only the points shown on the graph. In calculus, it's equivalent to what is called the "trapezoidal approximation". (It can be done with only three trapezoids, each with altitude 1.) Good work.

The actual area is exactly 15, as it turns out.
Good! I am only half a square off! I have almost finished the book I have been using...it is slow going when you do every problem, but if there is anything I've learned in life it is this: The more you practice, intelligently of course, the better. When I say intelligently I am thinking of an interview I saw the other day. A guitarist described her experience with focal dystonia.She is a professional guitarist, maybe 55 years old, who recently started having trouble with tremolo picking--this entails very rapid alternate picking with the first three fingers of the right hand (for a right handed guitarist). It is a relatively complicated, intense, and delicate procedure that involves a certain very small part of the brain. She used to sit watching television mindlessly practicing her tremolo for hours. Turns out this was her undoing. According to the diagnosis she received from physicians familiar with this sort of problem all that activity processed through that tiny patch of neurons had denatured them, burned them out in pop parlance. From great mastery she devolved to the clumsy maneuvering of an amateur--at least in the act of performing pieces with a lot of tremolo. I don't know if there is some form of focal dystonia mathematicians need to worry about or not. The world being as it is, I wouldn't be surprised.