Taylor polynomial of $\int_{0}^{x}\sin(t^2)dt$ I just learned about Taylor polynomials, and I am trying to estimate $\int_{0}^{1/2}\sin(x^2)dx$ using the 3rd degree Taylor polynomial of $F(x)=\int_{0}^{x}\sin(t^2)dt$ at $0$. I get the following:
$F'(x)=\sin(x^2)$,
$F''(x)=2x\cos(x^2)$,
$F^{(3)}(x)=2\cos(x^2)-4x^2\sin(x^2)$.
And so the estimate is:
$$\frac{F(0)}{0!}\cdot 1+\frac{F'(0)}{1!}\cdot \frac{1}{2}+\frac{F''(0)}{2!}\cdot \frac{1}{4}+\frac{F^{(3)}(0)}{3!}\cdot \frac{1}{8} = 0.$$
However this seems very weird (shouldn't it be closer to the actual value than $0$?), is my estimation correct?
 A: We first write down the Taylor expansion of $\sin t$ about $t=0$. This has undoubtedly already been done in your course. If it hasn't, calculation of the derivatives at $0$ is easy. We get
$$\sin t =t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\cdots.$$
Put $t=x^2$. We get
$$\sin(x^2)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}+\cdots.$$
Integrating term by term, we pbtain
$$F(x)=\frac{x^3}{3}- \frac{x^7}{7\cdot 3!}+\frac{x^{11}}{11\cdot 5!}-\frac{x^{15}}{15\cdot 7!}+\cdots.$$
The degree $3$ expansion is then simply $\dfrac{x^3}{3}$.  (Of course we need theorems to justify the substitution and the term by term integration.)
Let $x=1/2$. To estimate the error when we truncate at the $x^3/3$ term, it is in this case most efficient to note that our series is (for $x=1/2$) an Alternating Series. So the truncation error has absolute value less than $(1/2)^7/(7\cdot 3!)$.  The Lagrange form of the remainder, though useful for theoretical purposes, is often less useful when we want to produce good error estimates.  
A: We have 
$$f(x)=f(0)+xf'(0)+\frac{x^2}2f''(0)+\frac{x^3}{3!}f^{(3)}(0)+\int_0^x\frac{(x-t)^3}{3!}f^{(4)}(t)dt,$$
so 
\begin{align*}
\sin x^2&=x^2+\int_0^x\frac{(x-t)^3}{3!}(8\cos(t^2)(1-5t)+16t^4\sin(t^2))dt\\
&=x^2+\frac 43\int_0^x(x-t)^3(\cos(t^2)(1-5t)+2t^4\sin(t^2))dt,
\end{align*}
and for $x\geq 0$
\begin{align*}
\left|\int_0^x\sin t^2dt-\frac{x^3}3\right|&=\left|\int_0^x\sin t^2dt-\int_0^xt^2dt\right|\\
&\leq \frac 43\left|\int_0^x\int_0^t(t-s)^3(\cos(s^2)(1-5s)+2s^4\sin(s^2))dsdt\right|\\
&\leq \frac 43\int_0^x\int_0^t(t-s)^3(1+5s+2s^4)dsdt\\
&\leq \frac 43\int_0^x\int_0^tt^3(1+5s+2s^4)dsdt\\
&=\frac 13\left(\left[t^4\int_0^t(1+5s+2s^4)ds\right]_0^x-\int_0^xt^4(1+5t+2t^4)dt\right)\\
&=\frac 13x^4\left(x+\frac 52x^2+\frac 25x^5\right)-\frac 13\left(\frac{x^5}5+\frac 56x^6+\frac 29x^9\right)\\
&=\frac{x^5}3\left(\frac45+\frac 53x+\frac 25x^4-\frac 29x^4\right).
\end{align*}
