# Question #e5c2c

Nov 19, 2016

$F \left(x\right)$ is a strictly increasing function

#### Explanation:

If $F \left(x\right) = {\int}_{1}^{{x}^{2}} {e}^{- \sin \left(k\right)} \mathrm{dk}$ then

$\frac{d}{\mathrm{dx}} F \left(x\right) = {e}^{- \sin \left(x\right)} > 0$ so

$F \left(x\right)$ is a strictly increasing function

Dec 4, 2016

$F \left(x\right)$ is decreasing for $x \in \left(- \infty , 0\right)$
$F \left(x\right)$ has a minimum at $x = 0$
$F \left(x\right)$ is increasing for $x \in \left(0 , \infty\right)$

#### Explanation:

Suppose $G \left(x\right)$ is the antiderivative of ${e}^{- \sin \left(x\right)}$, i.e. $G ' \left(x\right) = {e}^{- \sin \left(x\right)}$. Then

$F ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{1}^{{x}^{2}} {e}^{- \sin \left(k\right)} \mathrm{dk}$

$= \frac{d}{\mathrm{dx}} \left(G \left({x}^{2}\right) - G \left(1\right)\right)$

$= \frac{d}{\mathrm{dx}} G \left({x}^{2}\right) - \frac{d}{\mathrm{dx}} G \left(1\right)$

$= 2 x G ' \left({x}^{2}\right)$

$= 2 x {e}^{- \sin \left({x}^{2}\right)}$

Then, as ${e}^{- \sin \left({x}^{2}\right)} > 0$ for all $x \in \mathbb{R}$, the sign of $F ' \left(x\right)$ will match the sign of $x$.

Because $F \left(x\right)$ is decreasing where $F ' \left(x\right) < 0$ and increasing where $F ' \left(x\right) > 0$, this gives us our final result:

$F \left(x\right)$ is decreasing for $x \in \left(- \infty , 0\right)$
$F \left(x\right)$ has a minimum at $x = 0$
$F \left(x\right)$ is increasing for $x \in \left(0 , \infty\right)$

This is the graph of $F \left(x\right)$: