For the function defined by $$F(x)=\begin{cases}\displaystyle\int_x^{2x}\sin t^2\,\mathrm dt,&x\neq0\\0,&x=0\end{cases}$$ analyze continuity and derivability at the origin. Is $F$ derivable at point $x_0=\sqrt{\pi/2}$? Justify the answer, and if possible, calculate $F'(x_0)$.
I have been told that I must use the Fundamental Theorem of Integral Calculus but I do not know how to apply it to this case.
For the function to be continuous at the origin, it must happen that $F(0)=\lim_{x\to0}F(x)$. We know that $F(0)=0$, and $$\lim_{x\to0}F(x)=\lim_{x\to0}\int_x^{2x}\sin t^2\,\mathrm dt\;{\bf\color{red}=}\int_0^{2\cdot0}\sin t^2\,\mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $\bf\color{red}=$.
To find the derivative at $x_0=0$ I tried the differentiate directly $F(x)$ but it is wrong, so I have been told that I must use the definition. So we have to find $$F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}{x-0}=\lim_{x\to0}\frac{\int_x^{2x}\sin t^2\,\mathrm dt}x.$$ Why we have to bound $\left|\sin t^2\right|\leq t^2$? How can we do that?
Finally, I do not know how to use the aforementioned theorem to justify that the function is derivable in $\sqrt{\pi/2}$. Using the definition again:
\begin{align*}
F'\left(\sqrt{\frac\pi2}\right)&=\lim_{x\to\sqrt{\frac\pi2}}\frac{F(x)-F\left(\sqrt{\frac\pi2}\right)}{x-\sqrt{\frac\pi2}}\\
&=\lim_{x\to\sqrt{\frac\pi2}}\frac{\int_x^{2x}\sin t^2\,\mathrm dt-\int_{\sqrt{\pi/2}}^{2\sqrt{\pi/2}}\sin t^2\,\mathrm dt}{x-\sqrt{\frac\pi2}}\\
&\leq\lim_{x\to\sqrt{\frac\pi2}}\frac{\int_x^{2x}t^2\,\mathrm dt-\int_{\sqrt{\pi/2}}^{2\sqrt{\pi/2}}t^2\,\mathrm dt}{x-\sqrt{\frac\pi2}}\\
&\underbrace=_{A=\sqrt{\pi/2}}\lim_{x\to A}\frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\\
&=\frac73\lim_{x\to A}\frac{x^3-A^3}{x-A}\\
&=\frac73\lim_{x\to A}\frac{(x-A)(x^2+Ax+A^2)}{x-A}\\
&=\frac73(A^2+A^2+A^2)\\
&=7A^2\\
&=\frac{7\pi}2,
\end{align*}
but it is wrong.
How can we solve the statement?
Thanks!
Best Answer
If you want to evaluate the limit:
$$\displaystyle\lim_{x\to 0}F(x)=\lim_{x\to 0}\int_{x}^{2x}\sin(t^2)dt$$
you can observe that $\forall x>0$ (the case $x<0$ is the same), $f(t)=\sin(t^2)$ is continuous in $[x,2x]$ so for the mean value theorem, exists $\xi_{x}\in (x,2x)$ such that
$$\int_{x}^{2x}\sin(t^2)dt=\sin(\xi_{x}^2)(2x-x)\implies F(x)=\sin(\xi_{x}^2)x$$
Now $\xi_{x}\to 0$ for $x\to 0^{+}$ so:
$$\lim_{x\to 0^{+}}F(x)=\lim_{x\to 0}\sin(\xi_{x}^2)x=[\sin(0)\cdot 0]=0$$
Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:
$$\lim_{x\to 0^{+}}\frac{F(x)-F(0)}{x-0}=\lim_{x\to 0}\frac{\sin(\xi_{x}^2)x}{x}=\lim_{x\to 0}\sin(\xi_{x}^2)=0$$
For $x_0=\sqrt{\frac{\pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.
$F'(x)=2\sin(4x^2)-\sin(x^2)\implies F'\left(\sqrt{\frac{\pi}{2}}\right)=2\sin(4\cdot\frac{\pi}{2})-\sin(4\cdot\frac{\pi}{2})=-1$