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

 

b a

I f x dx  

* The problem in numerical integration is the numerical evaluation of integral. * Geometrically, I is the area under the curve

• f f(x) between a and b.

Numerical Integration

Methods of solution

1- Trapezoidal Rule

 

1 1

2 h A f f  

x y b

f

 

y f x 

2

x

x

n

x

1

x

1

f

2

f

a

 

2 1 2

2 h A f f  

 

3 2 3

2 h A f f  

h h

   

1 2 1

2 .... 2

n n

h A f f f f f



         

b a h n  

Example

1 2 x

e dx 



1 0.2 5 h   

 

2x

f x e  x 0.6 0.2 0.4 0.8 1.0 f(x) 1 3.3201 1.4918 2.2255 4.9530 7.3890

   

0.2 1 7.3890 2 1.4918+2.2255+3.3201+4.9530 2 3.23698       

Approximate the value of , use n =5, then estimate the error.

1 2 x

e dx



Error Estimation

1 2 Exact x

e dx I  

1 2

2

x

e       

1 2

1 2

x

e     

2

1 1 2 e       3.19453 

.

3.23698

Trap

I 

. Exact Trap

Error I I  

3.19453 3.23698  

0.04245  

Example

2

1 x

e dx







1 0.1 10 h   

 

2

x

f x e  

0.1[(1 0.367879) (0.99005+0.960789+0.913931+0.852144 2 +0.778801+0.697676+0.612626+0.527292+0.444 2 858)]  

Approximate the value of , use n =10.

2

1 x

e dx





x 0.3 0.1 0.2 0.4 0.5 f(x) 1 0.913931 0.99005 0.960789 0.852144 0.778801 x 0.6 0.9 0.7 0.8 1.0 f(x) 0.697676 0.444858 0.612626 0.527292 0.367879

0.746211 

2- Simpson’s Rule

 

1 1 2

4 3 h A f f f   

 

2 2 3 4

4 3 h A f f f   

 

3 4 5 6

4 3 h A f f f   

   

1 3

4 .... 3

n

h A f f f f       

 

2 4

2 .... f f     

b a h n  

Example

1 0.1 10 h   

 

2x

f x e 

1 2 x

e dx 



x 0.3 0.1 0.2 0.4 0.5 f(x) 1 1.8221 1.2214 1.4918 2.2255 2.7183

     

0.1 1 7.3890 1.2214+1.8221+2.7183+4.0552+6.0496 3 1.4918+2.2255+3.3201+4.95 4 30 2       

x 0.6 0.9 0.7 0.8 1.0 f(x) 3.3201 6.0496 4.0552 4.9530 7.3890

Approximate the value of , use n=10, then estimate the error.

1 2 x

e dx



3.19454 

Example

1 2 Exact x

e dx I  

1 2

2

x

e       

1 2

1 2

x

e     

2

1 1 2 e       3.19453 

.

3.19454

Simp

I 

. Exact Simp

Error I I  

3.19453 3.19454   0.00001  

Example

2

1 x

e dx







1 0.1 10 h   

 

2

x

f x e  

0.1[(1 0.367879) (0.99005+0.913931+0.778801+0.612626+0.444858) 3 + (0.960789+0.852144+0.697676)+0.527 2 2 ] 4 29  

Approximate the value of , use n=10.

2

1 x

e dx





x 0.3 0.1 0.2 0.4 0.5 f(x) 1 0.913931 0.99005 0.960789 0.852144 0.778801 x 0.6 0.9 0.7 0.8 1.0 f(x) 0.697676 0.444858 0.612626 0.527292 0.367879

0.746825 