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Consider $f(x)=(1+x^2)^\frac{1}{3}$.

I've found the taylor polynomial of degree 2 centered around 0 to be

$(T_2 f)(x)=1+\frac{x^2}{3}$.

I am then asked to find

(1) a constant $C>0$ such that $|f(x)-T_2(x)|\leq C|x|^3$, for every $x\in[-1,1]$.

(2) Use (1) to find an approximation for the integral below and explain why the error is less than $C\frac{a^4}{4}$.

$I(a)=\int _{0}^a (1+x^2)^\frac{1}{3}\,dx$

But I'm not sure how to start (1) go about using the result in (2). While finding the talyor polynominal of degree 2 I also found $f^3(x)$ as I believe it can sometimes help later on, but it did not help me in this case.

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If you use the Lagrange form of the remainder, you have that $$\tag1 (1+x^2)^{1/3}=1+\frac{x^2}3-\frac89\,(1+\xi^2)^{-5/3}x^4, $$ with $\xi$ between $0$ and $x$ (and depending on $x$). So, using that $|x|\leq1$, $$\tag2 |f(x)-(T_2f)(x)|=\frac89\,\frac{x^4}{(1+\xi^2)^{5/3}}\leq\frac89\,x^4\leq\frac89\,|x|^3. $$ Using $(2)$, $$ \int_0^a[(1+x^2)^{1/3}-T_2f(x)]\,dx\leq\frac89\,\int_0^a|x|^3=\frac{8a^4}{36}=\frac{2a^4}{9}. $$ So, up to an error of $2a^4/9$, $$ \int_0^a(1+x^2)^{1/3}\,dx\simeq \int_0^a (1+\frac{x^2}3)\,dx=a+\frac{a^3}9 $$

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