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# Area under curve

Extracts from this document...

Introduction

In this investigation we will be estimating the area g(x) between X=0, X=1, this area is approximate by the sum of the two trapeziums’. Let one go 0→0.5 and the other 0.5→1. The area of a trapezium is ½ (a+b)h will be used to identify the total area throughout all the questions.

A₁ +A₂=(½ (3+3.25) x 0.5) + (½ (3.25+4) x 0.5)

= 1.56 + 1.81

= 3.37

= 0.2 here each trapezium will be 0.2 of each area, given that g (0) =3, g (0.2) = 3.04.... These values can be seen on the graph below. This area is approximate by the sum of the 5 trapeziums.

A₁ +A₂+A3+A4+A5 = (½ (3+3.04) x 0.2) + (½ (3.04+3.16) x 0.2) + (½ (3.16+3.36) x 0.2) +       (½ (3.36+3.64) x 0.2) + (½ (3.36+4) x 0.2)

= 0.60 + 0.62 + 0.65 + 0.70 + 0.76

= 3.34

With the help of GDC, we investigate what happens as the number of trapeziums increase:

A₁ + A₂ + A3 + A4 + A5 + A6 + A7 + A8 =(½ (3+3.015) x 0.125) + (½ (3.015+3.062) x 0.125) + (½ (3.062+3.14) x 0.125) + (½ (3.14+3.25) x 0.125) + (½ (3.25+3.390) x 0.125) +             (½ (3.390+3.562) x 0.125) + (½ (3.56+3.765) x 0.125) + (½ (3.765+3.4) x 0.125) =

= 0.375 + 0.379 + 0.389 + 0.399 + 0.415 + 0.434 + 0.457 + 0.485

= 3.333

A₁ + A₂ + A3 + A4 + A5 + A6 + A7 + A8 +A9 + A10 =(½ (3+3.01) x 0.1) + (½ (3.01+3.04) x 0.1) + (½ (3.04+3.09) x 0.1) + (½ (3.09+3.16) x 0.1) + (½ (3.16+3.25) x 0.1) +                                (½ (3.25+3.36) x 0.1) + (½ (3.36+3.49) x 0.1) + (½ (3.49+3.64) x 0.1) +                                   (½ (3.64+3.81) x 0.1) + (½ (3.81+4) x 0.1) =

= 0.3005 + 0.3025 + 0.3065 + 0.3125 + 0.3205 + 0.3305 + 0.3425 + 0.3565 + 0.3725 + 0.3905

= 3.335

(4 significant figures were used to demonstrate the difference between n= 8, n= 10)

A₁ + A₂ + A3 + A4 + A5 + A6 + A7 + A8 +A9 + A10 + A11 + A12 + A13 + A14 + A15 + A16 + A17 + A18 + A19 + A20=

(½ (3+3.002) x 0.05) + (½ (3.002+3.01) x 0.05) + (½ (3.01+3.02) x 0.05) +

(½ (3.02+3.04) x 0.05) + (½ (3.04+3.06) x 0.05) + (½ (3.06+3.09) x 0.1) +

(½ (3.09+3.12) x 0.05) + (½ (3.12+3.16) x 0.05) + (½ (3.16+3.20) x 0.05) +

(½ (3.20+3.25) x 0.05) + (½ (3.25+3.30) x 0.05) + (½ (3.30+3.36)

Middle

(g1 + g2) + (g2 + g3) +..................................+ (gn-1 + gn).

= ((g1 + g2) + (g2 + g3) + (g3 + g4) + ...........................................+ (gn-1 + gn)).

= (g0 + gn + 2g1 + 2g2 + 2g3 + ..........................................+2gn-1 +gn).

Total Area f(x) dx

dx (g0 + gn + 2(g1 + g2 + g3 + .......................................+gn-1)).  Equation [1]

Let c (g1 + g2 + g3 + ..........................................+gn-1).

C is the sum from n=1 using sigma summation notation.

C =

dx (g0 + gn + 2).

From the general case for g(x), we will be calculating the area under the curve for f (x) from x=b to x=a using n trapeziums.

From the diagram above:

D = b –a

D = nh

h =

Using the following notation:

f(a) = f0

f(a + h) = f1

f(a+2h) = f2

f(a+(n -1)h) = fn – 1

f(b) = fn

This follows a similar derivation as for g(x) from 1 to 0.

The following formula comes from equation [1] above, this can be used because the derivation for g(x) and f(x) is similar, the only difference being instead of saying b-a=1 we just use b-a in the expression.

dx (f0 + fn + 2f1 + 2f2 + ..........................................+2fn-1).

(f0 + fn + 2(f1 + f2 + ..........................................

Conclusion

GENERAL STATEMENT:

• The general statement accurately represents the area under the curve from b to a, if
• The curve doesn’t change rapidly with x, for example the following two curves demonstrate this for n=2.
• The larger the area you are examining the larger the value of n has to be in general
• [basically b-a/2n has to be a small value, in the examples above the value of 1/8 gave an error of about .5%, therefore it would be expected that if b-a/2n was of about this magnitude the trapezium rule would give an accurate fit of the area under the curve from b to a.
• Significant figures, because there is a difference in values, there wasn’t a set number of significant figures.
• The general statement doesn’t work, if b-a/2n is large
• For functions that change periodically, i.e. sin (x), and if you take trapeziums that are larger than one period of the function.
• If the function is negative in some areas, and these area are ignored by the edges of the trapeziums.

Overall the trapezium does provide a good fit to the area under the curve from b-a if the above conditions are met.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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