• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Shady Areas

Extracts from this document...

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

image09.png

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

...read more.

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.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Shady Areas. In this investigation you will attempt to find a rule to approximate ...

    This is because with more trapeziums it is possible to get a better approximation of the area since the line is going to a microscopic level, nearer to the original line. It gets nearer to the real value. Use the diagram below to find a general expression for the area

  2. Investigating the Ratios of Areas and Volumes around a Curve

    Because occurs, we cannot have n=-1 (for the moment). Graphically it can be seen that A would always equal 0, so the rule would still hold. As we cancel the terms in the final stage, a must not equal b.

  1. Investigating ratio of areas and volumes

    Now the conjecture must be tested for different arbitrary points. Consider the points between x = 0 and x = 2. Area A will be the area contained between the graph of y = xn and the x-axis between the points x = 0 and x= 2.

  2. The aim of this particular portfolio investigation is to find the area under a ...

    + g(x1)) h + 1/2 (g(x1) + g(x2)) h+....................+1/2 (g(xn-2) + g(xn-1)) h + 1/2 (g(xn-1) + g(xn)) h Adding all of the above together, we get the expression: Area = ( g(x0) + 2g(x1) + 2g(x2) + 2g(x3) +.................+ 2g(xn-1)

  1. Math IB SL Shady Areas Portfolio

    the area of a trapezoid is given as where is the height, and and are the lengths of the parallel sides. For the first trapezoid, the area is calculated to be: A1 And the second trapezoid: A2 And the total area being: Atotal A1 + A2 For the same function,

  2. Approximations of areas The following graph is a curve, the area of this ...

    Remember the width or base is the same for all the trapezoids. Area of a trapezoid (1) Area of Trapezoid (2) Area of Trapezoid (3) A(2)= A(2)= .604 .620 Area of trapezoid (4) Area of trapezoid (5) 00 Again we add all Trapezoids to retrieve the approximated area of the graph.

  1. type 1 maths portfolio trapezium rule

    area under the curve of the given function, we need to consider area for both trapeziums, This leads us to: Giving; It's good to compare the area got by estimation and area by using function to make notice of the limitations of approximation Therefore, Giving, And therefore arriving at or

  2. The investigation given asks for the attempt in finding a rule which allows us ...

    h=âxn=b-an From the previous question, the values in both first and last based are used twice.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work