# Determine the extrema of $\int^{x^2}_1\frac{\sin t}{2+e^t}\,dt$

Determine the local extrema of $$F(x)=\int^{x^2}_1\frac{\sin t}{2+e^t}\,dt$$.
Prove that $$|F(x)|\leq|x-1|$$ for all $$x$$.

$$F'(x)=\frac{2x\sin x}{2+e^x}\mbox{,}$$ so that the set of all points at which $$F$$ may attains its local extremum are $$n\pi$$, $$n\in\mathbb{Z}$$.

I don't know how to proceed any further.

• Have you, firstly, worked out the derivative of this function with respect to $x$? Commented Nov 22, 2021 at 8:42
• You should have some $x^2$ terms in your derivative. Commented Nov 22, 2021 at 8:49
• $F(1) = 0$ and $F'(x) < 1$ for all $x.$ So $F(x) \le \int_1^x t \ dt$ Commented Nov 22, 2021 at 8:52
• I see the second problem follows from the mean value theorem now.
– user912011
Commented Nov 22, 2021 at 8:56

The maybe more than one way to answer your question, however using the mean value theorem would be the most direct way to do it.

We have, $$|F'(x)|=\left|\dfrac{2x\sin(x^2)}{2+\exp(x^2)}\right|\leq\left|\dfrac{2x}{2+\exp(x^2)}\right|\leq\left| \dfrac{x}{1 + \frac{1}{2}\exp(x^2)}\right|\leq 1$$ The last inequality is valid for all $$x\in\mathbb{R}$$, because then we have $$2x\leq\exp(x^2)$$

Given that $$F(x)$$ is continuous on $$[1,x]$$ and derivable on $$]1,x[$$, we can conclude using the MVT: $$\left|\dfrac{F(x) - F(1)}{x-1} \right|\leq 1$$ Finall, given that $$F(1)=0$$ we have $$\left| F(x)\right| \leq \left| x-1 \right|, \forall x\in\mathbb{R}$$

About the local extrema: To find the local extrema of the function $$F$$, we need to solve the equation $$F'(x)=0$$.

$$F'(x)=0 \Longleftrightarrow \frac{2x\sin x^2}{2+\exp(x^2)}=0$$ Thus, $$x= 0,\quad \mbox{or }\ \sin(x^2)=0 \Leftrightarrow x^2=n\pi,\ n\in\mathbb{Z}$$

We can than conclude that the local extrema of the function are of the form $$\pm \sqrt{n\pi}, n \in \mathbb{N}$$

• How to determine the points at which $F$ attains it local maxima and minima?
– user912011
Commented Nov 22, 2021 at 9:27
• $3x^2=0$ when $x=0$ but $0$ is not an extrema of $x^3$
– user912011
Commented Nov 23, 2021 at 0:58
• Well, that's correct. However, that's not the case here, because the function $2x\sin(x^2)$ changes sign every time it crosses the $y=0$ line. Commented Nov 23, 2021 at 7:53