Simpson's rule of integration
- 1. Simpson’s Rule of Integration Dr. Varun Kumar Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 1 / 9
- 2. Outlines 1 Introduction to Simpson’s Rule 2 Mathematical Formulation 3 Example Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 2 / 9
- 3. Introduction to Simpson’s Rule Important points ⇒ Numerical integration means the numerical evaluation of integrals J = Z b a f (x)dx = F(b) − F(a), where F0 (x) = f (x) (1) a and b are given. f (x) is a function given analytically. J is the area under curve of f (x) between a and b. ⇒ As per Simpson’s rule We divide the interval of integration a 5 x 5 b into an even number of interval. Ex- n = 2m or sub-interval length h = b−a n Interval points are x0 = a, x1, x2, ......, x2m−1, x2m = b. Including end points, total 2m + 1 points are taken for consideration. Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 3 / 9
- 4. Continued– ⇒ We take and find the two sub-interval and approximate f (x) in the interval x0 5 x 5 x2 = x0 + 2h. ⇒ Applying the Lagrange’s polynomial p2(x) through (x0, f0), (x1, f1), (x2, f2) p2(x) = (x − x1)(x − x2) (x0 − x1)(x0 − x2) f0 + (x − x0)(x − x2) (x1 − x0)(x1 − x2) f1 + (x − x0)(x − x1) (x2 − x0)(x2 − x1) f2 (2) ⇒ The denominator terms are 2h2 , -h2 and 2h2 , respectively. ⇒ Let s = (x−x1) h then p2(x) = 1 2 s(s − 1)f0 − (s + 1)(s − 1)f1 + 1 2 (s + 1)sf2 (3) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 4 / 9
- 5. Geometrical Interpretation of Simpson’s Rule Figure: Simpson’s Rule Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 5 / 9
- 6. Algorithm of Simpson’s Rule Algorithm It computes the integral R b a f (x)dx Steps ⇒ INPUT: a, b, m, f0, f1, ....., f2m ⇒ Output: Approximate value ˜ J of J. ⇒ Compute s0 = f0 + f2m s1 = f1 + f3 + ... + f2m−1 s2 = f2 + f4 + ... + f2m−2 h = (b−a) 2m ˜ J = h 3 (s0 + 4s1 + 2s2) ⇒ Mathematical expression of numerical integration using Simpson’s Rule Z b a f (x)dx ≈ h 3 f0+4f1+2f2+4f3+...+2f2m−2+42f2m−1+f2m (4) Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 6 / 9
- 7. Example– Example Q Evaluate the following definite integral J with 2m = 10, where J = Z 1 0 e−x2 dx ⇒ Solve the above integral using Simpson’s rule and estimate the error. Ans Solution: According to question, a = 0, b = 1, 2m = 10 and h = 1−0 10 = 0.1 Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 7 / 9
- 8. Continued– ˜ J ≈ h 3 h 1.367879 + 4 × 3.740266 + 2 × 3.037901 i = 0.746825 Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 8 / 9
- 9. Integration accuracy Solve the standard integral and check its accuracy using rectangular and trapezoidal rule 1 R 1 0 x2dx 2 R 3 1 1 1+x dx 3 R π/2 0 cos x sin5 x dx 4 R 2 1 x2e2x dx 5 R 11 4 x3 1+x5 dx 6 R 1 0 x cos xdx Note: Students can also observe the effect of n on definite integral accuracy. Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 9 / 9