Taylor polynomial of an integral?

Question

How do you write the $(n+1)$th Taylor polynomial of an integral at $x_0$?

I.e. how do you expand $$T_{n+1, x_0} \left ( \int_{x_0}^xf(t)dt \right )$$

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The reason it's $(n+1)$th is because the Taylor expansion for just $ƒ(t)$ would be of order $n$.

Answer 1

It's really the Taylor expansion of the function $F(x) = \int_{x_0}^x f(t) \; dt.$ By Fundamental Theorem of Calculus, $F^{\prime}(x) = f(x)$. So the Taylor series start out $F(x_0)+F^{\prime}(x_0)(x-x_0) + F^{\prime\prime}(x_0)(x-x_0)^2/2 +\cdots.$ But (clearly) $F(x_0)=0$ and each $F^{(n)}=f^{(n-1)}$, so the series becomes $f(x_0)(x-x_0)+f^{\prime}(x_0)(x-x_0)^2/2 + \cdots.$

Alternatively, you could replace $f(t)$ by its Taylor series in the integrand and integrate term by term. (Valid as long as the series is absolutely convergent.)