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Integration

Integration. For a function f , The “integral of f from a to b ” is the area under the graph of the function. If f is continuous, then the area is well defined, as the common limit of upper and lower sums. The integral is denoted. graph of f ( x ). x. a. b.

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Integration

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  1. Integration For a function f, The “integral of f from a to b” is the area under the graph of the function. If f is continuous, then the area is well defined, as the common limit of upper and lower sums. The integral is denoted graph of f(x) x a b

  2. Numerical Integration: Trapezoid graph of f(x) approximate this region x a b graph of f(x) with this trapezoid x a b

  3. Composite Trapezoid graph of f(x) Apply “trapezoid” approximation to k subdivisions of [a b] x a b graph of f(x) “trapezoid” approximation with 3 subdivisions of [a b] x a b

  4. Composite Trapezoid graph of f(x) “trapezoid” approximation with 3 subdivisions of [a b] x a b h Let h=(b-a)/3. The sum of all the approximations is

  5. Simpson’s Rule: Suppose f(x)=x Let a = 2 and b = 8 ½(64-4) = 30 1/6(6)(2 + 4*5 +8) = 30

  6. Derivation of Simpson’s Rule: Suppose f(x)=x3

  7. Simpson’s Rule: Derivation Put all of those together, along with Hence: If f is any cubic polynomial, then This is the basis for Simpson’s rule.

  8. Simpson’s Rule For any function f, the Simpson’s approximation to is and in the middle Evaluate the function at the endpoints

  9. Composite Simpson’s Rule graph of f(x) Simpson on 3 subdivisions of [a b] h=(b-a)/3 x a b h Add them up. Total of 7 function evaluations.

  10. Numerical Integration: Ad-Hoc stopping criteria Pick a method (trapezoid, or Simpson’s). Set a stopping tolerance TOL. Pick k, an initial number of subdivisions Iterate as below • Apply composite method using k divisions • Apply composite method using 2k divisions • If answers are within TOL, stop, and return the 2k division answer • If answers are not within TOL, increase k and repeat. .

  11. Adaptive Stepsize Only use small h (the “stepsize”) where the convergence demands it. Recursive implementation is straightforward. function I = adr(fh,a,b,tol) Compute I1 using 1 subdivision Compute I2 using 2 subdivisions If the answers are within tol, I = I2; Else m = (a+b)/2; ILeft = adr(fh,a,m,tol/2); IRight = adr(fh,m,b,tol/2); I = ILeft + IRight; end

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