numerical integration and solving for limit
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hello everyone, this is my first question here.
I'm trying to find the value of x when t equals 1:0.1:10
So I don't know how to solve this problem, for example if the intervals were fixed and I wanted to find t, I would just use the commands
f(x)= (-1./(sqrt((1-x.^2)+(0.01/2)*(1-x.^4))))
a=1;
b=5;
I=int(f(x),a,b);
Can anybody help me?
how do i find the value of b when t is equal to 1:0.1:10.
Thanks for your help.
采纳的回答
Here is the maximum value you can insert for t:
syms x
f = 1/sqrt((1-x^2)+0.01/2*(1-x^4));
vpaintegral(f,x,-1,1)
And here is the curve of t against x.
You can cheat here: plot x against t and say you solved t = -integral ... for x. :-)
xstart = -1;
xend = 1;
xnum = linspace(xstart,xend,100);
for i=1:numel(xnum)
tnum(i) = double(vpaintegral(f,x,xnum(i),1));
end
plot(xnum,tnum)
4 个评论
Walter Roberson
2023-1-19
oh good point, the upper bound has to be in the range -1 to +1 (lower than the lower bound!) in order to not get a complex result.
Walter Roberson
2023-1-19
It looks to me as if -1 to +1 has an area of π so the t bounds would be 0 to π. It is the x bounds that must be -1 to 1 but the result of the integral can be larger than 1
Torsten
2023-1-19
Yes, I plotted x against the value of the integral.
PTK
2023-1-19
更多回答（1 个）
Walter Roberson
2023-1-19
0 个投票
your f has x^4 and x^2 but no other powers of x.
Do a change of variables X2=x^2 and integrate with respect to X2. You will get a closed form integral involving arcsin. Transform back from X2 to x. Now you can solve the equation. Just make sure you get the right limits of integration
2 个评论
PTK
2023-1-19
Walter Roberson
2023-1-20
The problem with this approach turns out to be that you would need x^2 to be negative to get at some of the values, which is a problem because that gets you into complex-valued x.
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