Taylor's Theorem: Lagrange Form for Approximating Error

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If the integral form of the remainder term gives the exact error, why do we use the Lagrange form of the remainder to approximate the error.
 
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All the common forms of the remainder Cauchy, Lagrange and Schlömilch; are not helpful in exact form. The exact answer is hard to compute. A simple approximation often is more useful than a complicated exact value.
 
Oh! that's so simple, thanks.
 
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