I f x dx a * Geometrically, I is the area under the curve - PowerPoint PPT Presentation

Numerical Integration * The problem in numerical integration is the numerical evaluation of integral. b     I f x dx a * Geometrically, I is the area under the curve of f(x) between a and b.

Methods of solution

   y f x 1- Trapezoidal Rule h     y A f f f 2 1 0 1 f 2 1 h     A f f f 2 1 2 0 2 h     A f f 3 2 3 2 h h x x x x x 1 0 2 n a b h             2 .... A  f f f f f   0 n 1 2 n 1 2  b a  h n

Example 1  2 x e dx Approximate the value of , use n =5, then estimate the error. 0    1 0  2 x   f x e h 0.2 5 x 0 0.2 0.4 0.6 0.8 1.0 2.2255 f(x) 1 1.4918 3.3201 4.9530 7.3890 1 0.2      e dx      2 x  1 7.3890 2 1.4918+2.2255+3.3201+4.9530  2 0  3.23698

Error Estimation   1 2 x I e dx Exact 0 1   2 x 1 1 e      2 x  e   2 0  2  0 1      2 3.19453 2 e 1    I 3.23698 Trap .   Error I I Exact Trap .   3.19453 3.23698   0.04245

Example 1   2 x e dx Approximate the value of , use n =10. 0  1 0   e  2    x h 0.1 f x 10 x 0 0.1 0.2 0.3 0.4 0.5 f(x) 1 0.99005 0.960789 0.913931 0.852144 0.778801 x 0.6 0.7 0.8 0.9 1.0 f(x) 0.697676 0.612626 0.527292 0.444858 0.367879 1 0.1[(1      2 x e dx 0.367879) 2 (0.99005+0.960789+0.913931+0.852144 2 0  +0.778801+0.697676+0.612626+0.527292+0.444 858)] 0.746211

2- Simpson ’ s Rule h      A f 4 f f 1 0 1 2 3 h      A f 4 f f 2 2 3 4 3 h      A f 4 f f 3 4 5 6 3 h           A f f 4 f f ....  0 n 1 3 3       2 .... f f  2 4  b a  h n

Example 1  2 x e dx Approximate the value of , use n=10, then estimate the error. 0    1 0  2 x   f x e h 0.1 10 x 0 0.1 0.2 0.3 0.4 0.5 f(x) 1 1.2214 1.4918 1.8221 2.2255 2.7183 x 0.6 0.7 0.8 0.9 1.0 4.9530 f(x) 3.3201 4.0552 6.0496 7.3890 1 0.1      e dx     2 x 1 7.3890 4 1.2214+1.8221+2.7183+4.0552+6.0496  3 0      3.19454 2 1.4918+2.2255+3.3201+4.95 30 

Example   1 2 x I e dx Exact 0 1   2 x 1 1 e    2 x    e   2 0  2  0 1      2 2 e 1 3.19453    I 3.19454 Simp .   Error I I Exact Simp .   3.19453 3.19454   0.00001

Example 1   2 x e dx Approximate the value of , use n=10. 0  1 0   e  2    x f x h 0.1 10 x 0 0.1 0.2 0.3 0.4 0.5 f(x) 1 0.99005 0.960789 0.913931 0.852144 0.778801 x 0.6 0.7 0.8 0.9 1.0 f(x) 0.697676 0.612626 0.527292 0.444858 0.367879 1    0.1[(1 2 x   e dx 0.367879) 4 (0.99005+0.913931+0.778801+0.612626+0.444858) 3 0  0.746825 + (0.960789+0.852144+0.697676)+0.527 2 29 2 ]