Is My Calculation Correct for Mid-Ordinate and Simpson's Rule?

  • Context: MHB 
  • Thread starter Thread starter peterianstaker
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on the accuracy of calculations for estimating the integral ∫(2,0) 1/(1+x^2) dx using the Mid-Ordinate rule and Simpson's rule with 4 strips. The Mid-Ordinate rule yields an estimate of 0.904, while Simpson's rule provides a more accurate estimate of approximately 1.105. The exact value of the integral is approximately 1.107, confirming that Simpson's rule is closer to the true value. The calculations utilize specific formulas for both numerical methods, demonstrating their application in estimating definite integrals.

PREREQUISITES
  • Understanding of definite integrals and their properties
  • Familiarity with numerical integration techniques, specifically Mid-Ordinate and Simpson's rules
  • Basic knowledge of calculus, including function evaluation and approximation methods
  • Ability to perform arithmetic operations with fractions and decimals
NEXT STEPS
  • Study the derivation and application of the Mid-Ordinate rule in numerical integration
  • Learn about Simpson's rule and its variations for different numbers of strips
  • Explore error analysis in numerical integration methods
  • Investigate other numerical methods for integration, such as Trapezoidal rule and Romberg integration
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are interested in numerical methods for solving integrals, as well as educators teaching calculus concepts.

peterianstaker
Messages
2
Reaction score
0
Can someone check my working to see if this is correct for both questions?

Estimate: ∫(2,0) 1/1+x^2 dx
i) Using the Mid-Ordinate rule with 4 strips
ii) Using Simpson's rule with 4 strips

i) h=b-a/n

= 2-0/4 = 0.5

x 0 0.5 1 1.5 2
y 1 0.8 0.5 0.308 0.2

=h(y1+y2+y3+...yn)
=0.5(0.8+0.5+0.308+0.2)
= 0.904 (3.d.p)

ii) = h/3(y0+4y1+2y2+4y3+y4)
= 0.5/3(1+4x0.8+2x0.5+4x0.3+4x0.308+0.2)
= 1.105 (3.d.p)

Any comments welcome!
 
Physics news on Phys.org
We are given:

$$f(x)=\frac{1}{x^2+1}$$

$$a=0,\,b=2,\,n=4$$

a) Midpoint Rule

$$M_4=\frac{1}{2}\left(\frac{1}{\frac{1}{16}+1}+\frac{1}{\frac{9}{16}+1}+\frac{1}{\frac{25}{16}+1}+\frac{1}{\frac{49}{16}+1}\right)=\frac{251168}{226525}\approx1.10878710959055$$

b) Simpson's Rule

$$S_4=\frac{1}{6}\left(\frac{1}{0+1}+4\frac{1}{\frac{1}{4}+1}+2\frac{1}{1+1}+4\frac{1}{\frac{9}{4}+1}+\frac{1}{4+1}\right)=\frac{431}{390}=1.1\overline{051282}$$

For comparison, the exact value of the given definite integral is:

$$I=\arctan(2)\approx1.10714871779409$$
 

Similar threads

MHB Calculating Integral with Simpson's Rule for Error < $0.5\cdot 10^{-3}$
  • · Replies 6 ·
Replies
6
Views
2K
Comp Sci Implementing Simpson's Rule in Fortran for Accurate Numerical Integration
  • · Replies 5 ·
Replies
5
Views
4K