I try to find a solution for
Try 1
syms x t
f=@(x)int(exp(-(t./x).^2),t,-1,1)-1.5;
fzero(f,[0.5,2])
Try 2
solve(int(exp(-(t./x).^2),t,-1,1)==1.5,x)
The first try does not work, the second work but I can enter an inverval How can I solve an equation given an interval?

 Risposta accettata

Torsten
Torsten il 6 Lug 2018
f=@(x)integral(@(t)exp(-(t/x).^2),-1,1)-1.5;
sol=fzero(f,[0.5,2])

Più risposte (1)

Walter Roberson
Walter Roberson il 6 Lug 2018

0 voti

vpasolve() permits you to enter ranges to search over.

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