to find area of parabola?

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RS
RS 2013 年 4 月 6 日
parabola going from three point (7.8 0.96), (8.25 0.99), (8.55 0.94), how can I compute area under the parabola?
x=[7.8 8.25 8.55];
y=[0.96 0.99 0.94];
p=polyfit(x,y,2);
f=polyval(x,p);
plot('x','y','o','x','f','-');

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bym
bym 2013 年 4 月 6 日
x=[7.8 8.25 8.55];
y=[0.96 0.99 0.94];
p=polyfit(x,y,2);
%f=polyval(x,p); x & p should be reversed
%plot('x','y','o','x','f','-'); read plot documentation
% method 1
t = linspace(x(1),x(3));
yhat = polyval(p,t);
a1 = trapz(t,yhat);
% method 2
f = @(v) p(1)*v.^2+p(2)*v+p(3);
a2 = quadgk(f,x(1),x(3));
a1,a2
a1 =
0.7344
a2 =
0.7344
  1 件のコメント
RS
RS 2013 年 4 月 6 日
Thank you!! I have another problem
x1=7.8;
x2=8.5;
y1=0.96;
y2=0.94;
p1 = polyfit([x1 x2], [y1 y2], 2);
b1= polyval(p1,1);
m1=polyval(p1,2)-b1;
x3=8.25;
x4=8.25;
y3=0;
y4=.99;
p2 = polyfit([x3 x4], [y3 y4], 2);
b2 = polyval(p2, 1);
m2 = polyval(p2, 2) - b2;
I got x value = -1.2867; from which co-ordinate this value corresponds to? Actually I want to compute intersection of two line with respect to x=[7.8 8.25 8.5]; y=[0.96 0.99 0.94]; over which those two lines are plotted?
x=[7.8 8.25 8.5];
y=[0.96 0.99 0.94];
p=polyfit(x,y,2);
f=polyval(p,x);
plot(x,y,'o',x,f,'-');
hold on;
line([7.8 8.5],[0.96 0.94]);
hold on;
line([8.25 8.25],[0 0.99]);

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