how find exact value q for integral on partition interval?

i want find exact value for q(i)

let f(x,y)

x = s : h : p ;

i = 2 : n-1 ;

q ( i ) = integral ( f( x , y ( i ) ) , x( i-1 ), x( i+1 ) ).

for example:

q=integral(x)dx on interval 0:2:8

output

q1=2,

q2=6,

q3=10,

q4=14. please i attached it

[John D'Errico]

HUH? Totally unclear. Please explain. CLEARLY. SLOWLY. The question as is makes no sense.

[work wolf]

i edit and explain it.

Answer 1

[John BG]

This question is simple, but not as simple as it looks at first sight.

the function x you have defined

x=[0:2:8]
has a reference vector 

nx=[1:5]

your attempt to integrate beyond 5 does not work unless the reference vector of the function covers such interval.

To get the integration results q1=2 q2=6 q3=10 q4=14 the function you should integrate could be

x=2*ceil([1:8]/2)

and reference vector

xn=0:1:8

the problem here is that MATLAB interpolates the following way

stair01

despite showing an apparently staired stem plot

stair02

and the integration results along the intervals are biased with constant 2:

quad(x,0,2)
 =
          4.00

quad(x,2,4)
 =
          8.00

quad(x,4,6)
 =
         12.00

quad(x,6,8)
 =
         16.00

to solve the excessive interpolation that MATLAB performs by default, here linear interpolation is excessive, you may want to do the following:

1.- increase resolution

nx2=0:0.01:8

2.- be aware that if you define the following function

function y=func1(n)
% generate stair with even integers only
  n2=floor(n)
  if n2<1
    y=0
    return
  end
  if(~mod(n2,2))   % even
    y=n2
    else                % odd
        y=n2+1
    end
end

it seems to work but it does not. If you try

func1([1:1:8])
n2 =
     1     2     3     4     5     6     7     8
y =
     2     3     4     5     6     7     8     9
ans =
     2     3     4     5     6     7     8     9

despite one by one

func1(0.3)
=
0

func1(1)
=
2

func1(2.1)
=
2

func(3.1)
=
2

func(3.9)
=
2

func(4.6)
=
4
..

func1 seems to work fine with scalars but not with vectors. You need custom defined functions to work correctly with vectors to pass the handle to the integrating function quad or integral.

if you try quad(@func1,a,b) with the previous function func1 on the sought intervals the results are wrong.

4.- So, to generate the values of the function to integrate correctly you have to introduce the following loop in the function that calculates the stair:

for k=1:1:length(nx) 
   y(k)=func1(nx(k)); 
end

now if you plot y you get the right function:

just did it, use the following function

function yn=func1(n)

n2=floor(n)

yn=zeros(1,length(n))

for k=1:1:length(n)

if(~mod(n2(k),2))   % even
    yn(k)=n2(k)
    else
        yn(k)=n2(k)+1
end

end

end

now you get the right results:

n=[0:.1:8]

quad(@func1,0,2)
 =
   2.000008674804688

quad(@func1,2,4)
=
   6.000008674804688

quad(@func1,4,6)
 =
  10.000008674804688

quad(@func1,6,8)
=
 14.000008674804688

If you find this answer of any help solving your question,

please click on the thumbs-up vote link,

thanks in advance

John

[e-mail]

[work wolf]

really , thank you so much , you are always helping.