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  • Level: GCSE
  • Subject: Maths
  • Word count: 1966

Area Under a Straight Line Graph - Calculate the area under a straight line graph.

Extracts from this document...

Introduction

Area Under a Straight Line Graph

Task:

Calculate the area under a straight line graph.  Investigate.

Plan:

I intend to start off with simple diagrams eg y = x, y = x + 1, y = x + 2 etc.

I will then calculate the areas and put them into a table.  I can then use the table and the diagrams to help me find a generalisation.  I can then draw more complex diagrams by changing the gradient and the intercept.  I can then find generalisations with several variables.  I then will test and explain these generalisations.

image00.png

Basic Diagrams:image01.png

image12.pngimage13.pngimage14.pngimage16.pngimage15.png

Equation

Area of ‘triangle’

Area of ‘rectangle’

Total Area

y = x

(10 x 10) / 2 = 50

-

50 units2

y = x + 1

50

1 x 10 = 10

6o units2

Y= x + 2

50

2 x 10 = 20

70 units2

y = x + 3

50

3 x 10 = 30

80 units2

y = x + 4

50

4 x 10 = 40

90 units2

y = x + 5

50

5 x 10 = 50

100 units2

y = x + 10

50

10 x 10 = 100

150 units2

Generalisation:

x2 / 2 + xc  

This generalisation I found for when the gradient is 1.

X2 / 2  stands for the area of the triangular section.  This is because x2  would give you the area if it was a square as x is the length of the side.  It is divided by 2 as it is a triangle.

Xc is the area of the rectangular section.

Triangular section.

                                                     Rectangular section

Therefore the generalisation stands for the total area!

Test of 1st generalisation:

Taking y = x + 2 …

Total Area = x2 / 2 + xc

                      102 / 2 + (10 x 2)

                     (100 / 2) + 20

                     50 + 20

                     70 units2

Taking y = x + 10 …

Total Area = x2 / 2 + xc

                      102 / 2 + ( 10 x 10)

                      (100 / 50) + 100  

                       50 + 100

                       150 units2

...read more.

Middle

3 x 10 = 30

130 units2

y= 2x +4

100

4 x 10 = 40

140 units2

y = 2x + 5

100

5 x 10 = 50

150 units2

y =2x + 10

100

10 x 10 = 100

200 units2

2nd Generalisation:

x2m / 2 + xc

This generalisation was found when the gradient is 2.  

x2m / 2 stands for the triangular section.  The change from the last generalisation is the m.  The m means that the section is multiplied by the gradient to get the right area of the whole because the gradient is more than 1!  You can then divide by 2 and you have the correct area for the triangular section.  xc is for the same reasons as in the 1st gradient.

Test of 2nd generalisation:

Taking y = 2x + 1…

Total Area = x2m / 2 + xc

                     102(2) / 2 + 10(1)

                     100(2) / 2 + 10

                     200 / 2 + 10

                     100 + 10

                     110 units2

Taking y = 2x + 5…

Total Area = x2m / 2 + xc

                     102(2) / 2 + 10(5)

                     100(2) / 2 + 50

                     200 / 2 + 50

                     100 + 50

                   150 units2

Although I found this generalisation to fit when the gradient is 2 I think that it will work with other gradients as well as the m is a variable.  To find out if I am right I am going to do some more tests to see whether I will get the right answers or not.  To do this however I have to look at more complex diagrams.  I am however just going to sample a couple to see whether they will work.  There isn’t the need now for as many as before as I have already found the generalisation I wish to test.

More complex Diagram… continued:image06.pngimage07.png

Equation

Area of ‘triangle’

Area of ‘rectangle’

Total Area

y = 3x + 3

(10 x 30)/2 = 150

3 x 10 = 30

180 units2

y = 3x +5

150

5 x 10 = 50

200 units2

y= 4x + 2

(10 x 40)/2 = 200

2 x 10 = 20

220 units2

y = 4x + 10

200

10 x 10 =100

300 units2

image08.pngimage09.png

Test of 2

...read more.

Conclusion

image11.pngDraw the step-graph which starts by dropping and then going horizontally

across the width of the vertical strips (take 5m as the example again!).
          The area under this lower step-graph is less than the area under the curve itself.

image11.png        The area under the curve itself is between the 2 areas found from the upper

and lower step-graphs.  We can then take the mean of these 2 values as a reasonable approximation to find the approximate area.

By looking at the two methods I believe that the Trapezium Rule is a lot better!  It is quicker, easier to use, and is a lot more accurate to the true area!

Conclusion:

The Trapezium Rule is the best way to find the area under a curved line graph.  I also discovered that the areas found are a lot more accurate when the lowest value possible is used for the widths.  For instance if the width of 10 goes equally across the graph, then it is better to halve the widths and use 5 as you get a more accurate answer!

With more time:

If I had been given more time, then I would have introduced some A-level work, such as calculus and differentiation.  This is because I planned to research these and teach them to me to find the area under curved line graphs but I didn’t have enough time!  I also would have included more examples of the Trapezium Rule – in practice and in working examples; as well as using detailed diagrams and examples of the Step Method.

...read more.

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