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POLYVAL
polyval p ( x ) = 3 x 2 + 2 x + 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 ( 3 x 4 − 4 x 2 + 1 0 x − 2 5 ) d x . Create a vector to represent the polynomial integrand 3 x 4 − 4 x 2 + 1 0 x − 2 5 . The x 3 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]))
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