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Another statement of Taylor's Theorem which gives a different form for the remainder is the following: Let [math]f[/math] be a function with continuous [math](n+1)\mathrm{st''[/math] derivative at every point of the interval [math][a,b][/math]. Then

[[math]] f(b) = f(a) + f^\prime(a)(b-a) + \cdots + \frac1{n!} f^{(n)} (a)(b-a)^n [[/math]]

[[math]] + \int_a^b \frac{(b-t)^n}{n!} f^{(n+1)} (t) \; dt . [[/math]]

}

• lab{9.8.12a} Using integration by parts, show that
  [[math]] \int_a^b \frac{(b-t)^n}{n!} f^{(n+1)} (t) \; dt [[/math]]
  [[math]] = - \frac1{n!} f^{(n)} (a)(b-a)^n + \int_a^b \frac{(b-t)^{n-1}}{(n-1)!} f^{(n)} (t) \; dt . [[/math]]
• Using induction on [math]n[/math] and the result of part \ref{ex9.8.12a}, prove the above form of Taylor's Theorem in which the remainder appears as an integral.