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∫ π 2
Pb. 4.24 2 dx
0 1+ cos ( )x

Example 4.6
∫ x
Plot the value of the indefinite integral 0 fx dx() as a function of x, where f(x)
is the function sin(x) over the interval [0, π].

Solution: We solve this problem for the general function f(x) by noting that:

∫ 0 x f x dx() ≈ ∫ 0 x−∆ x f x dx() + f x( − ∆ x + ∆ x / )2 ∆ x (4.5)

where we are dividing the x-interval into subintervals and discretizing x to
correspond to the coordinates of the boundaries of these subintervals. An
array {x } represents these discrete points, and the above equation is then
k
reduced to a difference equation:

Integral(x ) = Integral(x ) + f(Shifted(x ))∆x (4.6)
k–1
k
k–1
where
Shifted(x ) = x + ∆x/2 (4.7)
k–1
k–1
and the initial condition is Integral(x ) = 0.
1
The above algorithm can then be programmed, for the above specific func-
tion, as follows:

a=0;
b=pi;
dx=0.001;
x=a:dx:b-dx;
N=length(x);
xshift=x+dx/2;
yshift=sin(xshift);
Int=zeros(1,N+1);
Int(1)=0;
for k=2:N+1
Int(k)=Int(k-1)+yshift(k-1)*dx;

© 2001 by CRC Press LLC