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It is easy to identify: $\large f(x)=A(x-1729)(x-1730)(x-1731)+x-1725$ (See Polynomial Interpolation )

Hence, $\large{\displaystyle\int _{ 1729 }^{ 1731 }f(x)dx \\ \begin{aligned}&=\displaystyle\int _{ 1729 }^{ 1731 }(A(x-1729)(x-1730)(x-1731)+x-1725)\,dx\\&\\&\\&\color{#D61F06}{\text{Substituting x-1730=t such that dx=dt}}\\& \\&=\int_{-1}^{1} (A(t+1)(t)(t-1)+t+5)\,dt\\&\\&= \int_{-1}^{1}(\color{#20A900}{At^3-At+t}+5)\,dt\\&\\\Large&=\color{#20A900}{0}+10\\&\\&=\huge\boxed{10}\end{aligned}}$

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$**\int_{-1}^{1} (\color{#20A900}{At^3-At+t})\,dt=0$

since $\color{#20A900}{At^3-At+t}$ is a odd function by using- if $f(x)$ is an odd function, then $\displaystyle \int _{ -a }^{ a }{ f(x)dx }=0.$

We can use Simpson's rule , which gives the exact value of the integral for polynomials of degree $\le 3$ . Therefore,

$\displaystyle\int _{ 1729 }^{ 1731 }{ f(x)dx=\frac { 1731-1729 }{ 6 } \left[ f(1731)+4f\left( \frac { 1731+1729 }{ 2 } \right) +f(1729) \right] } =10$