HP71B Integral Questions

[Albert Chan]

(02-07-2020 01:19 PM)Wes Loewer Wrote:  Yes, those are the points that you would expect using midpoints. These are the same points used on the 50g.

Since trapezoids and midpoints are using the same points, I tried to tease out the algorithm, using f(x) = x*log(1+x)

10 DEF FNF(X)=X*LOGP1(X)
20 DEF FNG(U)=1.5*(1-U*U)*FNF(U*(3-U*U)/2)
30 M1=2*FNG(0)
40 M2=FNG(-.5)+FNG(.5)
50 M4=.5*(FNG(-.75)+FNG(-.25)+FNG(.25)+FNG(.75))
60 T1=0
70 T2=(T1+M1)/2
80 T4=(T2+M2)/2
90 T8=(T4+M4)/2
100 DISP "MIDPOINTS=";(M1-20*M2+64*M4)/45
110 DISP "TRAPEZOID=";(-T1+84*T2-1344*T4+4096*T8)/2835
120 DISP "INTEGRAL =";INTEGRAL(-1,1,1,FNF(IVAR))

>RUN
MIDPOINTS= 1.02693666365
TRAPEZOID= .978017069767
INTEGRAL7= .97801706977

For this case, T₁ = 0 is a valid assumption.

>FNG(-.9999), FNG(+.9999)
5.40429438145E-3     2.07933751591E-4

For a detail comparison between these 2 methods: https://math.stackexchange.com/questions...int/674350