First HP: (IF) Arc integral Leght returning another integral
Hello!
I'm trying to adjust the length of the arc of the function y (x) = 2.5 + 9 * xˆ2 - 0.5xˆ4, between the points x = 0 and x = 4.
The calculation to do is to solve the inegral from 0 to 4 of dx sqrt (1 +(18*x-2*xˆ3) ˆ2)
The question is:
When trying to solve this particular integral, the calculator returns me another full, recursively, as if he cannot completely solve the first integral.
Someone knows how to set up the calculator to be able to solve this kind of integral?
My CASE is currently:
Angle: radians
Digital format: Standard
Simplify: no
Exact [v]
Use sqrt [v]
Main [v]
Evaluation of recursive: 10
Recursive replacement: 2
Recursive function: 20
Epsilon: 1E-12
Probability: 1E-15
Newton: 80
P.S: Wolfram Alpha solves this kind of integral.
Hello
Fortunately, there is no facility to do here!
Basically, you ask the calculator to solve with a perfect, finished form which I am not sure exists in this case. It works for a while and when he cannot make further progress, returns then the internal form modified "as close to the perfect result, accurate" that he can. In this case, it is a square root value and a modified integral.
Press SHIFT + ENTER to run the INPUT function shifted to an approximation to this outcome and the decimal answer comes out. This scribble = sign is the mathematical symbol for "closer of".
Note that if you include one. in the initial entry, the approximate would immediately take out as the CASE hit this number apprixmate and then continues a form of "decimal approximation.
SQRT (1.+(18*x-2*xˆ3) ˆ2) <-note the. After that the 1 indicates the calculator "is not perfect, idea of 1 but rather an approximate number of 1. It's a pretty big difference in mathematics.
If you put a 1. in your original entry, he would immediately go out the decimal. However, just using the SHIFT ENTER key as needed is usually the easiest way to do.
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