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We know area of it is 30.

because 1/2bh... now how do u express it in integrals?

And how do you find the value of y at which the triangle is divided into 2 equal sections?

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- Thread starter PrudensOptimus
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- #1

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We know area of it is 30.

because 1/2bh... now how do u express it in integrals?

And how do you find the value of y at which the triangle is divided into 2 equal sections?

- #2

Integral

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y = 5x/12

between x=0 and x=12

[inte] 5x/12 dx = 5x

evaluate at the upper limit (lower limit yields zero)

=5(144)/24= 30

I am not quite sure what you mean by the second part, put suddenly I feel as if I have done your homework, so.... I'll let you puzzel out the second part, haveing seen how I did the first part.

- #3

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what is 5x/12? bh/2, isn't that the area of the triangle?

the 2nd part is what I wanted to know a long time for a real life problem. I have a garden, with a shape like a triangle. I was thinking of adding a wall to it so that it makes 2 sections, 1 for vegetables and 1 for others, but both have the same area.

The problem I am thinking might relate to calculus, that's what I heard, ... But I don't know how to use calculus to find where should i start (0,y) to add the wall...

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ok, i see why you said find area under 5/12x...

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Integral

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A = 5/24 x

You have an expression for the Area, given x, what you want to do is find x where the Area = 15

15 = 5/24 x

x = sqrt (72)= 6 sqrt(2)

This is the distance from the smallest angle along the long (not hypotenuse) side.

If that not the side you want the wall from, simply rewrite the equation for the line, only with the approiate slope you orient your garden correctly.

(Does that make sense?) :)

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can you explain what the "dx" part mean in the integral?

My understanding for integral is not very good, only know how to solve a few of them, but I would like to know what the dx parts mean so that I can make more use of it.

And we took the integral of the line y = 5/12x,... that gives us the area. So that means the derivative of area is a line that borders the area??

also, what is the "x" in the integrated equation?

thanks.

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- #8

HallsofIvy

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One way of defining "integral" is that ∫

One way of approximating that is to divide the x axis into many pieces, each of length &DELTA; x. Construct a rectangle on each piece with height f(x

In your case, a 90 degree triangle with legs of length 5 and 12 DOESN'T HAVE any graph or equation connected with it so you CAN'T find the area by an integral!

Well, you can, IF you place a coordinate system on it so that you can think of it as a graph. Staring at a right triangle for a very long time (I'm kinda slow!), it starts to dawn on me that the two legs look a little like x and y axes so I consider setting up a coordinate system with the right angle at (0,0), the x-axis along one leg, and the y-axis along the other. Now I notice that the hypotenuse is (part of) a line passing through the points (0,12) and (5,0) (if you set up your coordinate system so that it passes through (12,0) and (0,5), Okay, that'll work too.).

The line passing through both (0,12) and (5,0) has slope (12-0)/(0-5)= -12/5 and so its equation is y= -(12/5)(x-0)+ 12 or y= -(12/5)x+12. (Integral set up his coordinate system so that leg with length 12 was the x-axis and the leg with length 5 was the y-axis. He got y= -(5/12)(x-12) as the equation of the line {though he mistakenly wrote it as y= -(5/12)x!!). Okay, that'll work too.)

The area is, according to the theory, ∫

30+ C- C= 30. Exactly the same as (1/2)(5)(12)!!!!

- #9

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I never understood the LRAM or whatever it is called approximation method, is that important?

Never looked at Rieman sums either:\

However many way you approached the triangle problem, I thought it can be modeled as integral of 5/12x dx because the area is between the line y=5/12x and y=0.

Thus integral of 5/12x - 0 = integral of 5/12x dx.

And would someone please answer my questions prior to his response pls.

- #10

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I choose the simpler line. Place the origin, not at that right angle but at the either of the other angles so the hypotenuse passes through the origin. Now the hypotenuse IS the line we wish to find the area below. Since the line passes through the origin (by design) we simply use the length of the other 2 sides as defining the slope, using rise over run (y/x). The line is given by y=mx you now simply integrate the equation of the line with limits of the origin to the length of the side on the x axis.

Thus the integral yields a slightly different, but equivalent, to the standard expression for the area of the triangle, it is different because it incorporates our knowledge of the relationship between the 2 sides.

in general the area = mx^{2}/2

where m is the slope of the hypotenuse which must pass through the origin.

There is no mistake! The line I used is y=5/12 x.

Simplify,simplify,simplify!

Thus the integral yields a slightly different, but equivalent, to the standard expression for the area of the triangle, it is different because it incorporates our knowledge of the relationship between the 2 sides.

in general the area = mx

where m is the slope of the hypotenuse which must pass through the origin.

{though he mistakenly wrote it as y= -(5/12)x!!). Okay, that'll work too.)

There is no mistake! The line I used is y=5/12 x.

Simplify,simplify,simplify!

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- #11

Integral

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Originally posted by PrudensOptimus

can you explain what the "dx" part mean in the integral?

My understanding for integral is not very good, only know how to solve a few of them, but I would like to know what the dx parts mean so that I can make more use of it.

And we took the integral of the line y = 5/12x,... that gives us the area. So that means the derivative of area is a line that borders the area??

also, what is the "x" in the integrated equation?

thanks.

OPPs, missed this,

The dx is simply notation, it must follow the integral symbol, view it as a form of a period, it blocks off the function to be integrated.

The x in my expression

A = mx

Edit: I missed the square in my exprssion for area.

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but the width(height) is only 5 meters total.

- #13

Integral

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One http://home.comcast.net/~rossgr1/Math/Garden.pdf is worth a thousand words.

I have a feeling you are going to say, "But that is not how I wanted the wall to run!"

Look back at my first post I said

You never bothered to clear that up, you only spec was to split the garden in half. The dotted line does that.

I have a feeling you are going to say, "But that is not how I wanted the wall to run!"

Look back at my first post I said

I am not quite sure what you mean by the second part

You never bothered to clear that up, you only spec was to split the garden in half. The dotted line does that.

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- #14

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ok, can i plug that back into bh/12 =15, where b is 6sqrt(2) and find h?

- #15

Integral

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Yes if you mean bh/2=15,Or you could plug it into 5x/12 where x=6 sqrt(2), or you could read it off my graph ~3.5

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- #17

HallsofIvy

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Suppose the problem is, to use your example, to find the area between the x-axis (y=0) and the graph of y= sin(x) between x=0 and x= π . The upper boundary of that is y= sin(x), the lower boundary is y=0 so the integrand is sin(x)- 0= sin(x). The integral is ∫

∫

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