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Shady Areas

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Introduction

Shady Areas

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.

...read more.

Middle

image03.png

=            x  ½

=            x

Y1 =                    When n = 8 using the general formulaimage05.png

image05.pngimage06.png

image07.png

image43.pngimage08.png

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Therefore            

                             = 1/8 x 15.8394

                             = 1.9799image44.pngimage11.png

image12.png

Y2  =

image12.pngimage05.png

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

image13.png

image45.png

image14.png

image15.png

Y3 =

image05.png

image17.png

Integration of these functions to find the

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Conclusion

160;           = (9x/√(x³ + 9)) =  ?

image05.png

image20.png

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

image19.pngimage18.pngimage22.png

     =    x^4 – (23/3)x³ + 20x² - 18x

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

     = 0 – (-4.667)

     = 4.67

image23.pngimage26.pngimage27.pngimage25.pngimage28.png

image32.pngimage32.pngimage32.pngimage32.pngimage32.pngimage29.pngimage33.pngimage30.pngimage05.pngimage31.png

image37.pngimage05.pngimage35.pngimage36.pngimage34.pngimage35.png

              (9x/√(x³ + 9))

image38.pngimage39.pngimage37.pngimage05.pngimage40.pngimage41.png

              (4x³ - 23x² + 40x - 18)image37.png

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.

...read more.

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