The integrand has poles at -2.5 and 3.3, so let's look for a solution that is in the interval (-2.5, 3.3).
Also, all the messages "rat replaced ..." are distracting, so let's set ratprint to false. Finally, let's
do the calculation in several steps.
(%i1) e : integrate(1/(2.9*(3.3-u)-(3.3-u)^2/2), u, 2.26, x), ratprint : false;
"Is "50*x-113" positive, negative, or zero?"neg;
"Is "2*x+5" positive, negative, or zero?"pos;
(%o1) (10*log(2*x+5))/29-(10*log(33-10*x))/29+0.030486537015191
The function solve is unable to solve e = 0.1.
(%i2) solve(e=0.1,x);
rat: replaced -0.06951346298481 by -7107353/102244266 = -0.06951346298481
(%o2) [log(33-10*x)=(1022442660*log(2*x+5)-206113237)/1022442660]
So let's try a numerical solution. To guess a solution, look at a plot. To draw the graph, use
plot2d(%o1,[x,-2.2,3]). The function find_root gives a root:
(%i3) find_root(%o1=0.1,x,2.2,2.5);
(%o3) 2.485018482603471
--Barton
________________________________________
Hi everyone,
First of all I am italian so sorry for my English...
Anyway, my problem is quite simple (I think) but I'm getting mad on it!
I have to solve a quite simple equation where the unknown value is the
second bound of the integral.
I wrote this:
solve(0.1=integrate(1/(2.9*(3.3-u)-(3.3-u)^2/2), u, 2.26, x), x);
and this is the result:
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced 2.26 by 113/50 = 2.26
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced 3.3 by 33/10 = 3.3
rat: replaced 2.9 by 29/10 = 2.9
rat: replaced 3.3 by 33/10 = 3.3
rat: replaced -2.26 by -113/50 = -2.26
"Is "50*x-113" positive, negative, or zero?"positive;
rat: replaced -4.76 by -119/25 = -4.76
rat: replaced -2.26 by -113/50 = -2.26
"Is "10*x-52/5" positive, negative, or zero?"positive;
rat: replaced 1.04 by 26/25 = 1.04
rat: replaced -2.9 by -29/10 = -2.9
rat: replaced 3.3 by 33/10 = 3.3
rat: replaced 2.9 by 29/10 = 2.9
rat: replaced 3.3 by 33/10 = 3.3
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced 2.26 by 113/50 = 2.26
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced 2.26 by 113/50 = 2.26
rat: replaced -2.26 by -113/50 = -2.26
rat: replaced -10.4 by -52/5 = -10.4
rat: replaced 10.0 by 10/1 = 10.0
rat: replaced 9.52 by 238/25 = 9.52
rat: replaced 2.0 by 2/1 = 2.0
Principal Value
rat: replaced 0.069513462984809 by 1203/17306 = 0.069513463538657
rat: replaced -1.08330781158269 by -3537/3265 = -1.0833078101072
rat: replaced 0.3448275862069 by 10/29 = 0.3448275862069
rat: replaced -0.3448275862069 by -10/29 = -0.3448275862069
rat: replaced -0.3448275862069 by -10/29 = -0.3448275862069
rat: replaced 0.3448275862069 by 10/29 = 0.3448275862069
(%o38)
[log(2*x+5)=(565040900*log(10*x-33)+565040900*log(-1)-1775128338*%i+113906055)/565040900]
I answered positive because I know that the result should be 2.48 but in
the end it comes out with imaginary values...
I also tried to set logabs to true and solveradcan to true but nothing
good happened.
How can I solve that equation?
Thank you very much.
Michele Minelli
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