Hi, Chuck --

Quote:

... But on page 15-11 of the 35s manual, it says "SOLVE cannot call a routine that contains an Integral(FN) instruction" and visa-versa (or am I missing something here?)

You're not missing anything. The ability to run SOLVE within INTEG or vice-versa was not provided on the HP-32S in 1988, and that shortcoming has been carried forward on all its successors -- the HP-32SII, the HP-33s, and the HP-35s.

From my only HP Forum article:

http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/articles.cgi?read=556

Quote:

The SOLVE and INTEG functions were introduced on the HP-34C in 1979, and later directly ported to the HP-15C in 1982. For the HP-41C/CV/CX, the same functions were adapted with only slight modification in the Advantage Pac introduced in 1985. Consequently, the protocol for using SOLVE and INTEG on these models is almost identical ... Moreover, the program may include a SOLVE or INTEG instruction, thus allowing INTEG to be executed within SOLVE, or vice versa. This can be useful for solving a limit of integration, or a parameter of an integrand.
...

*
Neither the HP-32S, the HP-32SII, the HP-33s, nor the HP-35s provide the capability to execute INTEG within SOLVE, or vice versa. The HP-42S, however, retained that capability of the original implementation.*

6 = Int[sqrt(1 + 4t^2),t,0,x)

Calculation of the arc length of a function is a standard calculus problem, e.g.: sqrt[(dx)^2 + (dy)^2] = sqrt[1 + (dy/dx)^2].dx

This function is apparently parameterized as

x = t; y = t^{2}, so

sqrt[(dx/dt)^2 + (dy/dt)^2].dt = sqrt[1 + (2t)^2].dt

The integrand will be modestly larger than 2t. Integrating 2t.dt from 0 to x such that the integral will be equal to 6.0 yields x = sqrt(6) = 2.449. The actual solution will be lower than that.

Here's a program and result returned by my 1982-made LED HP-34C, which is executable without modification on an HP-15C, and as adapted (quite minimally) for the HP-41 with Advantage ROM:

Program:
HP-15C/34C HP-41 with Advantage

LBL A LBL "AA"

0 0

x<>y x<>y

INTEG B "AB"

6 INTEG

- 6

RTN -

LBL B RTN

x^{2} LBL "AB"

4 x^{2}

* 4

1 *

+ 1

SQRT +

RTN SQRT

RTN

Execution:

FIX 3 FIX 3

2 2

ENTER ENTER

3 3

SOLVE A "AA"

SOLVE

(Answer: 2.305419905)

FIX 4 FIX 4

2.305 2.305

ENTER ENTER

2.307 2.307

SOLVE A "AA"

SOLVE

(Answer: 2.306097460)

This is a good application of INTEG within SOLVE, the mechanics of which I've done before. However, I've yet to encounter or come up with a realistic and practical application of SOLVE within INTEG.

-- KS

*Edited: 9 Feb 2008, 1:46 a.m. after one or more responses were posted*