For a function defined by parts study continuity, and differentiability at two points

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!

• Would it help if you write $\int_x^{2x}=\int_0^{2x}-\int_0^x$? – A.Γ. Jan 3 at 23:59
• @A.Γ. probably. Can we separate it into two integrals because the limits of integrations are continuous? – manooooh Jan 4 at 0:01
• It is because the integral is additive. Or if you wish $\int_a^b=F(b)-F(a)$ where $F(y)$ is antiderivative, which is e.g. $\int_0^y$. – A.Γ. Jan 4 at 0:05
• @A.Γ. how do you know that $0\in[x,2x]$ for all $x\neq0$? Because for example if $x\in(0,\infty)$ then $0\not\in[x,2x]$. – manooooh Jan 4 at 0:29
• One does not need $0\in[x,2x]$. Integral is additive and if $f$ is Riemann integrable on some closed interval containing $a, b, c$ then $$\int_{a} ^{b} f(x) \, dx+\int_{b} ^{c} f(x) \, dx=\int_{a} ^{c} f(x) \, dx$$ irrespective of the linear order of $a, b, c$. This assumes that we have by definition $\int_{a} ^{a}f(x)\,dx=0$ and $\int_{b} ^{a} f(x) \, dx=-\int_{a} ^{b} f(x) \, dx$. – Paramanand Singh Jan 4 at 7:23

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 $$F(x) =2\int_{0}^{x}\sin (4z^2)\,dz-\int_{0}^{x}\sin t^2\,dt$$ and by FTC we can see that $$F$$ is continuous and differentiable everywhere with derivative $$F'(x) =2\sin (4x^2)-\sin x^2$$ for all $$x\in\mathbb {R}$$.

For reference I mention FTC explicitly :

Fundamental Theorem of Calculus Part 1: Let the function $$f:[a, b] \to\mathbb {R}$$ be Riemann integrable on $$[a, b]$$. Then the function $$F:[a, b] \to\mathbb {R}$$ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ is continuous on $$[a, b]$$ and if $$f$$ is continuous at some point $$c\in[a, b]$$ then $$F$$ is differentiable at $$c$$ with derivative $$F'(c) =f(c)$$.

Using the above theorem it can be proved that if a function $$f:\mathbb {R} \to\mathbb {R}$$ is Riemann integrable on every bounded and closed interval then the function $$F:\mathbb {R} \to\mathbb {R}$$ defined by $$F(x) =\int_{a} ^{x} f(t) \, dt$$ for some $$a\in\mathbb {R}$$ is continuous everywhere and if $$f$$ is continuous at some point $$c\in \mathbb {R}$$ then $$F$$ is differentiable at $$c$$ with $$F'(c) =f(c)$$.

• @manooooh: if you carefully note the FTC you will see that it deals with integrals where upper limit of integral is $x$ and lower limit is a constant. The substitution in first integral is done to change upper limit $2x$ to $x$. The second integral is already having $x$ as upper limit. If we apply substitution in second integral also then the upper limit changes from $x$ to $x/2$ and FTC can't be applied directly. Continued in next comment. – Paramanand Singh Jan 5 at 1:34
• @manooooh: Using FTC and chain rule for derivatives it can be proved that if $$F(x) =\int_{a} ^{g(x)} f(t) \, dt$$ then $F'(x) =f(g(x)) g'(x)$. This handles the case when the upper limit of integral is not $x$ but rather some complicated function $g(x)$. Regarding your second doubt, you can check using this formula that the answer remains same even if we apply substitution in second integral. – Paramanand Singh Jan 5 at 1:39
• @manooooh: you need to revisit substitution in definite integrals in your textbook and observe very carefully the given examples. During substitution in a definite integral the function as well as the limit of integral change according to the substitution used. Thus if we use $z=t/2, t=2z,dt=2\,dz$ in second integral the limits $0$ and $x$ change to $0$ and $x/2$ to give $$\int_{0}^{x}\sin t^2\,dt=2\int_{0}^{x/2}\sin (4z^2)\,dz$$ – Paramanand Singh Jan 5 at 4:32
• @manooooh: the limits of integral indicate the range of values being taken by the variable of integration. Thus in $\int_{0}^{x}\sin t^2\,dt$ the variable $t$ varies from $0$ to $x$. If $z=t/2$ then $z$ should vary from $0$ to $x/2$ (as $z$ is half of $t$). – Paramanand Singh Jan 5 at 4:37
• @manooooh: irrespective of how proficient you are in mathematics, your comments indicate that you have a sincere desire to learn. That's what matters here and nothing else. +1 for your question. – Paramanand Singh Jan 5 at 4:40

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.

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

• Thanks for the answer! How do you know that $F(x)=f(\xi_{x})(2x-x)$? From Mean Value Theorem, we have that exists $\xi_{x}\in (x,2x)$ such that $F(x)=\frac{F(2x)-F(x)}{2x-x}$. – manooooh Jan 4 at 0:49
• I applied the mean value theorem for the function $f(t)=\sin(t^2)$ over the intervall $[x,2x]$. So exists $\xi_x\in(x,2x)$ such that $\int_{x}^{2x}f(t)dt=f(\xi_x)(2x-x)$. – Ixion Jan 4 at 0:52
• I do not know how do you get $f(\xi_x)(2x-x)$. Since $f(x)$ is continuous at $[x,2x]$ and $f(x)$ is differentiable at $(x,2x)$ then exists $\xi_x\in(x,2x)$ such that $\require{cancel}f'(\xi_x)=\dfrac{f(2x)-f(x)}{2x-x}=\dfrac{\sin(4x^2)-\sin(2x)}x\cancel\implies f(\xi_x)(2x-x)$. – manooooh Jan 4 at 5:31
• The mean value theorem states that: if $f(t)$ is continuous in $[a,b]$ then it exists $\xi\in (a,b)$ such that: $$\int_{a}^{b}f(t)dt=f(\xi)(b-a)$$ In this case $f(t)=\sin(t^2), \ a=x$ and $b=2x.$ – Ixion Jan 4 at 21:06

Since $$\frac {\sin\, x} x\to1$$as $$x \to 0$$ we can find $$\delta >0$$ such that $$\frac 1 2 t^{2} \leq\sin(t^{2})\leq 2t^{2}$$ for $$|t| <\delta$$. This gives $$\frac 7 6 x^{3} \leq F(x) \leq \frac {14} 3x^{3}$$ for $$0 and it follows easily from the definition that the right hand derivative of $$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)=2\sin(4x^{2})-\sin(x^{2})$$. At the given point $$x_0$$ the derivative is $$-1$$.

• Thanks for the answer! Why do you use "Since $\frac {\sin\, x} x\to1$ as $x\to0$ (...)"? – manooooh Jan 4 at 6:35
• @manooooh it makes it easy to see that $\frac {F(x)} x \to 0$ as $x\to 0$. I can use just the definition of derivative instead of using MVT, etc. – Kavi Rama Murthy Jan 4 at 6:40
• Ok. Could you give me the guidelines on how to use the derivative definition at a point, please? From $$F'(0)=\lim_{x\to0}\frac{F(x)-F(0)}{x-0}=\lim_{x\to0}\frac{\int_x^{2x}\sin t^2\,\mathrm dt}x$$ I should use the fact that $\left|\sin t^2\right|\leq t^2$ so now the limit becomes $\lim_{x\to0}\left|\frac{\int_x^{2x}t^2\,\mathrm dt}x\right|$? – manooooh Jan 4 at 6:44
• @manooooh Now calculate he integral of $t^{2}$ from $x$ to $2x$. You will get $7x^{3} /3$. You now see that the limit is $0$, right? – Kavi Rama Murthy Jan 4 at 7:17
• @manooooh You said you have been asked to use the Fundamental Theorem of Calculus. That is what I have done in my answer. DO not try to find the derivative at $x_0$ using the definition of derivative. That is messy! – Kavi Rama Murthy Jan 4 at 7:30