Chapter 6: Techniques of Integration
Section 6.7: Numeric Methods
Example 6.7.7
Use the Trapezoid rule to approximate the area under the curve determined by the following data points.
k
0
1
2
3
4
5
6
7
8
9
10
xk
−1
−0.8
−0.6
−0.4
−0.2
0.2
0.4
0.6
0.8
fk
13
7.523
4.679
3.528
3.436
4.971
6.175
7.440
8.516
Table 6.7.7(a) Data points determining a curve
Solution
Mathematical Solution
Figure 6.7.7(a) shows the eleven data points and the piecewise linear curve connecting them.
The Trapezoid rule, with n=10 and h=1/5 gives the area A as
A
=h2f0+f10+2∑k=19fk
=1/5213+9+250.268
=110122.536
=12.2536
use plots in module() local X,Y,S,q,f,p1,p2,p3,k; q:=-2*x^5+2*x^4-4*x^3+5*x^2+4*x+4; for k from 0 to 10 do f[k] := evalf(eval(q,x=-1+k/5),4); end do; X:=[seq(k/5,k=-5..5)]: Y:=[seq(f[k],k=0..11)]: S:=[seq([X[k],Y[k]],k=1..11)]: p1:=plot(X,Y,view=[-1..1,0..13]): p2:=plot(S,style=point,symbol=solidcircle,symbolsize=15,color=green): p3:=display(p1,p2,labels=[x,y]); print(p3); end module: end use:
Figure 6.7.7(a) Piecewise linear curve
Of course, the most tedious part of the calculation is entering the data!
Maple Solution
Enter the data
Form a list of the eleven function values in Table 6.7.7(a).
Context Panel: Assign to a Name≻f
13,7.523,4.679,3.528,3.436,4,4.971,6.175,7.440,8.516,9→assign to a namef
Apply the Trapezoid rule
Because lists are enumerated from 1, the eleven data points must be referenced as fk,k=1,…,11.
1/52f1+f11+2∑k=210fk = 12.25360000
The use of a list for the function values is a simplification that is counterbalanced by the need to shift the indices of the data points upward by 1.
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