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Extracts from this document...

Introduction

I shall investigate different ways of finding a working rule to approximate the area under a curve using trapeziums.  The area under the curve represents area between the f(x) and x values under the curve in the specified area.  Therefore, through integration to find the area, integration of the integrated area will find the volume.

Consider the function   g(x) = x² + 3

g(x) = x² + 3  (see Graph 1)

The area under this curve from x=0 to x=1 is approximated by the sum of the area of two trapeziums.

The approximation can be discovered by working out the area of the square in the trapezium and then by working out the triangle in the trapezium.

Middle

=            x  ½

=            x

Y1 =                    When n = 8 using the general formula

Therefore

= 1/8 x 15.8394

= 1.9799

Y2  =

=  1/8  x  [y0 + 2y1 +2y2 + 2y3 + 2y4 + 2y5 + 2y6 + 2y7 + y8]

Y3 =

Integration of these functions to find the

Conclusion

160;           = (9x/√(x³ + 9)) =  ?

Y3 =  m        (4x³ - 23x² + 40x - 18)

=    x^4 – (23/3)x³ + 20x² - 18x

=  (81 – 207 + 180 – 54) - (1 - (23/3) +20 – 18)

= 0 – (-4.667)

= 4.67

(9x/√(x³ + 9))

(4x³ - 23x² + 40x - 18)

The first approximation was 15.6% off from the integrated answer while the third approximation was 7.7% off from the integrated answer.  The third would be closer as the curve has a minima and maxima and, as the trapeziums overestimate with maxima and underestimates with curves with a minima, and so reduces the percentage error as it cancels it out.

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

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