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       #Post#: 69--------------------------------------------------
       p1
   DIR By: zacker93
       Date: January 9, 2014, 5:47 am
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       Como coņo se plotea la segunda funcion????
       >> x=0:0.01:4;
       >> y=sinh(exp(-x));
       >> plot(x,y)
       >> grid
       >> hold on
       >> y2=0.5*(1-sqrt(4*x-1*x.^2));
       >> y3=0.5*(1+sqrt(4*x-1*x.^2));
       >> plot(x,y2)
       >> plot(x,y3)
       apartado b)
       >> p1=fzero('0.5*(1+sqrt(4*x-1*x.^2))-sinh(exp(-x))',[0 1])
       >> p2=fzero('0.5*(1-sqrt(4*x-1*x.^2))-sinh(exp(-x))',[3 4])
       >> A1=quad('sinh(exp(-x))-0.5*(1+sqrt(4*x-1*x.^2))',0,p1)
       >> A2=quad('sinh(exp(-x))-0.5*(1-sqrt(4*x-1*x.^2))',1,p2)
       #Post#: 70--------------------------------------------------
       Re: p1
   DIR By: antay
       Date: January 9, 2014, 5:55 am
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       Pero solo te sale la del senh(exp(x))... y el dibujo de las
       dos??Vamos a por el B!!!
       #Post#: 71--------------------------------------------------
       Re: p1
   DIR By: zacker93
       Date: January 9, 2014, 5:55 am
       ---------------------------------------------------------
       tio es lo que estoy preguntando
       #Post#: 72--------------------------------------------------
       Re: p1
   DIR By: antay
       Date: January 9, 2014, 6:17 am
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       El corte de la derecha he puesto:
       >> p=fzero('sinh(exp(-x))-(1/2*sqrt(4*x-x.^2)+1)',[0,2])
       y me da p=0.0208
       el area me da
       >> A=quad('-sinh(exp(-x))+1/2*(1-sqrt(4*x-x.^2))',p2,p1)
       A =
       2.2029
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