Respuesta :

When we divide the interval from 1 to 5 into 4 subintervals, each has a width of 1. Simpson's rule has us evaluate the integral as

... integral = (1/3)(f(1) +4f(2) +2f(3) +4f(4) +f(5)) = (1/3)(10 +4·25 +2·46 +4·73 +106)

... integral = (1/3)(600) = 200 . . . . . by Simpson's rule

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The integral evaluates to

[tex]\displaystyle \int_{1}^{5}{\left(3x^2+6x+1\right)}\,dx=\left[x^3+3x^2+x\right]\limits_{1}^{5}=(5^3-1^3)+3(5^2-1^2)+(5-1)\\=124+3*24+4=200[/tex]

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Simpson's rule uses a quadratic interpolation, so evaluates quadratics exactly.

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