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Compute a quadrature of

$\int_c^d\int_a^b f(x,y)dxdy$

using the Simpson rule and estimate the error.

So the Simpson rule says

$S(f) = (b-a)/6(f(a)+4f((a+b)/2) +f(b))$

So i get

$\int_c^d(b-a)/6(f(a)+4f((a+b)/2) +f(b))dy$

Is that even correct? How do I go on?

  • 0
    Are you looking for the general theory or is there a specific $f(x, y)$ and limits? – 2017-01-30
  • 0
    General theory, there is a part exercise with a specific function, but I think I'll be able to do this then – 2017-01-30
  • 0
    Try: http://jmahaffy.sdsu.edu/courses/s10/math541/lectures/pdf/week09/lecture.pdf and http://www.math.usm.edu/lambers/mat460/fall09/lecture32.pdf – 2017-01-30

1 Answers 1

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Something to get you started. Define an operator: $$ I_x(f) := \int_a^b f(x) \, dx $$ We can see that $I_x(f + g) = I_x(f) + I_x(g)$ and if $\alpha$ does not depend on $x$, $I_x(\alpha f) = \alpha I_x(f)$. In other words the operator is linear.

If it acts on $f(x, y)$ it returns a function of $y$ only.

By analogy define another operator $$ I_y(f) := \int_c^d f(y) \, dy $$

Applying $I_x$ and $I_y$ consequently we get a double integral: $$ I_y(I_x(f)) := \int_c^d \left[ \int_a^b f(x, y) \, dx \right] dy . $$

Now define a "Simpson's rule" operator: $$ S_x(f) := (f(a) + 4 f((a+b)/2) + f(b)) (b - a)/6 $$ It is also linear and if it acts on $f(x, y)$ it "kills" dependency on $x$. The same for $S_y$: $$ S_y(f) := (f(c) + 4 f((c+d)/2) + f(d)) (d - c)/6 $$

By analogy with integral let compute $S_y(S_x(f(x, y)))$ and call it Simpson's rule for double integral. Apply $S_x$ to f(x, y): $$ S_x(f(x, y)) = (f(a, y) + 4 f((a+b)/2, y) + f(b, y)) (b - a)/6 $$ and apply $S_y$ to the result using linearity $$ S_y(S_x(f(x, y)) = (S_y(f(a, y)) + 4 S_y(f((a+b)/2, y)) + S_y(f(b, y))) (b - a)/6 $$ expand all three terms with $S_y$ $$ S_y(S_x(f(x, y)) = \\ \left(16f\left(\frac{b+a}{2},\frac{d+c}{2}\right)+4f\left(\frac{b+a}{2},d\right)+4f\left(\frac{b+a}{2},c\right)+4f\left(b,\frac{d+c}{2}\right) +f\left(b,d\right)+f\left(b,c\right)+4f\left(a,\frac{d+c}{2}\right)+f\left(a,d\right)+f\left(a,c\right)\right) \frac{\left(b-a\right)\left(d-c\right)}{36} $$