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For the function defined by $$F(x)=begin{cases}displaystyleint_x^{2x}sin t^2,mathrm dt,&xneq0\0,&x=0end{cases}$$ 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 $F(0)=lim_{xto0}F(x)$. We know that $F(0)=0$, and $$lim_{xto0}F(x)=lim_{xto0}int_x^{2x}sin t^2,mathrm dt;{bfcolor{red}=}int_0^{2cdot0}sin t^2,mathrm dt=0,$$ so the statement holds, but here I do now how to justify the $bfcolor{red}=$.

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 $$F'(0)=lim_{xto0}frac{F(x)-F(0)}{x-0}=lim_{xto0}frac{int_x^{2x}sin t^2,mathrm dt}x.$$ Why we have to bound $left|sin t^2right|leq t^2$? 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{fracpi2}right)&=lim_{xtosqrt{fracpi2}}frac{F(x)-Fleft(sqrt{fracpi2}right)}{x-sqrt{fracpi2}}\
&=lim_{xtosqrt{fracpi2}}frac{int_x^{2x}sin t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}sin t^2,mathrm dt}{x-sqrt{fracpi2}}\
&leqlim_{xtosqrt{fracpi2}}frac{int_x^{2x}t^2,mathrm dt-int_{sqrt{pi/2}}^{2sqrt{pi/2}}t^2,mathrm dt}{x-sqrt{fracpi2}}\
&underbrace=_{A=sqrt{pi/2}}lim_{xto A}frac{1/3((2x)^3-x^3)-1/3((2A)^3-(A^3))}{x-A}\
&=frac73lim_{xto A}frac{x^3-A^3}{x-A}\
&=frac73lim_{xto A}frac{(x-A)(x^2+Ax+A^2)}{x-A}\
&=frac73(A^2+A^2+A^2)\
&=7A^2\
&=frac{7pi}2,
end{align*}

but it is wrong.

How can we solve the statement?

Thanks!

If you want to evaluate the limit:

$$displaystylelim_{xto 0}F(x)=lim_{xto 0}int_{x}^{2x}sin(t^2)dt$$

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 $xto 0^{+}$ so:

$$lim_{xto 0^{+}}F(x)=lim_{xto 0}sin(xi_{x}^2)x=[sin(0)cdot 0]=0$$

Note that this argument can be used to show that $F(x)$ is derivable for $x=0$, infact:

$$lim_{xto 0^{+}}frac{F(x)-F(0)}{x-0}=lim_{xto 0}frac{sin(xi_{x}^2)x}{x}=lim_{xto 0}sin(xi_{x}^2)=0$$

For $x_0=sqrt{frac{pi}{2}}$, the derivative of $F(x)$ can be found using the Fundamental Theorem of Integral Calculus.

$F'(x)=2sin(4x^2)-sin(x^2)implies F'left(sqrt{frac{pi}{2}}right)=2sin(4cdotfrac{pi}{2})-sin(4cdotfrac{pi}{2})=-1$

You just need to use the fundamental theorem of calculus. Since the integrand $sin(t^2)$ is continuous everywhere we can write $$F(x) =int_{0}^{2x}sin t^2,dt-int_{0}^{x}sin t^2,dt$$ Use substitution $z=t/2$ in first integral on right to get $$F(x) =2int_{0}^{x}sin (4z^2),dz-int_{0}^{x}sin t^2,dt$$ and by FTC we can see that $F$ is continuous and differentiable everywhere with derivative $$F'(x) =2sin (4x^2)-sin x^2$$ for all $xinmathbb {R} $.

Since $frac {sin, x} xto1 $as $x to 0$ we can find $delta >0$ such that $frac 1 2 t^{2} leqsin(t^{2})leq 2t^{2}$ for $|t| <delta$. This gives $frac 7 6 x^{3} leq F(x) leq frac {14} 3x^{3}$ for $0<x<sqrt {delta}$ and it follows easily from the definition that the right hand derivative of $F$ at $0$ is $0$. Make the substitution $s=-t$ to see that the left hand derivative is also $0$. Hence $F'(0)=0$. For $x>0$ we have $F(x)=int_0^{2x}sin(t^{2}), dt -int_0^{x}sin(t^{2}), dt$ from which it follows (by Fundamental Theorem of Calculus) that $F'(x)=2sin(4x^{2})-sin(x^{2})$. At the given point $x_0$ the derivative is $-1$.