Derivation of Simpson\'s Rule PDF

Title Derivation of Simpson\'s Rule
Author Hi Hi
Course Math
Institution Singapore University of Technology and Design
Pages 2
File Size 78.7 KB
File Type PDF
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Simpson’s Rule Simpson’s rule is a numerical method that approximates the value of a definite integral by using quadratic polynomials. Let’s first derive a formula for the area under a parabola of equation y = ax2 + bx + c passing through the three points: (−h, y0 ), (0, y1 ), (h, y2 ). y

y = ax2 + bx + c

(0, y1 )

(−h, y0 )

(h, y2 )

y1

y2

y0 ∆x 0

−h

A= =

Z

h

x

h

(ax2 + bx + c) dx −h



ax3 bx2 + + cx 3 2

h   

−h

2ah3 = + 2ch 3  h = 2ah2 + 6c 3

Since the points (−h, y0 ), (0, y1 ), (h, y2 ) are on the parabola, they satisfy y = ax2 + bx + c. Therefore, y0 = ah2 − bh + c y1 = c y2 = ah2 + bh + c Observe that y0 + 4y1 + y2 = (ah2 − bh + c) + 4c + (ah2 + bh + c) = 2ah2 + 6c. Therefore, the area under the parabola is A=

∆x h (y0 + 4y1 + y2 ) = (y0 + 4y1 + y2 ) . 3 3

Gilles Cazelais. Typeset with LATEX on April 23, 2008.

We consider the definite integral Z

b

f (x) dx.

a

We assume that f (x) is continuous on [a, b] and we divide [a, b] into an even number n of subintervals of equal length b−a ∆x = n using the n + 1 points x0 = a,

x1 = a + ∆x,

x2 = a + 2∆x,

...,

xn = a + n∆x = b.

We can compute the value of f (x) at these points. y0 = f (x0 ),

y1 = f (x1 ),

y2 = f (x2 ),

...,

yn = f (xn ).

y

y4

y0 y1

a = x0 x1

y2

x2

yn

y3

x3

yn−2

x4

···

∆x

···

yn−1

xn−2 xn−1 xn = b

x

We can estimate the integral by adding the areas under the parabolic arcs through three successive points. Z b ∆x ∆x ∆x (y2 + 4y3 + y4 ) + · · · + (y0 + 4y1 + y2 ) + (yn−2 + 4yn−1 + yn ) f (x) dx ≈ 3 3 3 a By simplifying, we obtain Simpson’s rule formula. Z

b a

f (x) dx ≈

∆x (y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 4yn−1 + yn ) 3

Example. Use Simpson’s rule with n = 6 to estimate Z

1

Solution. For n = 6, we have ∆x = x √ y = 1 + x3

4−1 6

1 √ 2

4p

1 + x3 dx.

= 0.5. We compute the values of y0 , y1 , y2 , . . . , y6 . 1.5 √ 4.375

2 3

2.5 √ 16.625

3 √ 28

3.5 √ 43.875

4 √ 65

Therefore, Z

1

4p

1 + x3 dx ≈

√ √ √ √ √  0.5 √ 2 + 4 4.375 + 2(3) + 4 16.625 + 2 28 + 4 43.875 + 65 3

≈ 12.871...


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