polyval

 p(x)=3x2+2x+1 at the points x=5,7,9

p = [3 2 1];
x = [5 7 9];
y = polyval(p,x)

y = 1×3

    86   162   262




I=3−1(3x4−4x2+10x−25)dx.




Create a vector to represent the polynomial integrand 3x4−4x2+10x−25. The x3 term is absent and thus has a coefficient of 0.



p = [3 0 -4 10 -25];




Use polyint to integrate the polynomial using a constant of integration equal to 0.



q = polyint(p)






q = 1×6

    0.6000         0   -1.3333    5.0000  -25.0000         0






Find the value of the integral by evaluating q at the limits of integration.



a = -1;
b = 3;
I = diff(polyval(q,[a b]))