# Second mean-value theorem for integrals, Apostol

Let $f''(x)$ be continous and non zero on $[a,b]$ and if $f'(x)\geq m >0,~ \forall x,\in[a,b]$, prove that:

$$\left|\int _a^b \sin(f(x))\,dx\right|\leq \frac{4}{m}$$

Now, I need to use this theorem:

$$\int_a^bf(x)g(x)\,dx=f(a)\int_a^cg(x)\,dx+f(b)\int_c^bg(x)\,dx$$

There is also a hint that I should multiply the integrand by $\frac{f'(x)}{f'(x)}$.

I would be pleased if somebody could help me with directions, of how I should proceed?

Obviously, I should try to leave $f'(x)$ or $\frac{1}{f'(x)}$ in the integrand right? But if I try that I don't seem to get far. Maybe there is an intermediate step? Thanks!

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## 2 Answers

$\def\abs#1{\left|#1\right|}$We have for some $c \in (a,b)$: \begin{align*} \abs{\int_a^b \sin f(x) \, dx} &= \abs{\int_a^b f'(x)\sin f(x) \cdot \frac 1{f'(x)}\, dx}\\ &\le \frac 1{f'(a)}\abs{\int_a^c f'(x)\sin f(x)\, dx} + \frac 1{f'(b)}\abs{\int_c^b f'(x)\sin f(x)\, dx}\\ &\le \frac 1m \abs{\int_{f(a)}^{f(c)} \sin u \, du} + \frac 1m \abs{\int_{f(c)}^{f(b)}\sin u\, du} \end{align*} Now note, that for any interval $[\alpha, \beta]$ we have writing $\beta = \alpha + 2k\pi + \gamma$ where $\gamma < 2\pi$ and $k \in \mathbb N$ \begin{align*} \abs{\int_\alpha^\beta \sin u \, du} &= \abs{\int_0^{\gamma}\sin u \, du}\\ &\le \int_0^{\pi} \abs{\sin u}\, du\\ &= 2. \end{align*} Pluging this into the above gives the desired estimate.

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brilliant! wish I could come up with these ideas :/ – Sarunas Nov 19 '12 at 9:44

A less clever, although valid, proof:

First, observe that \begin{aligned} |\cos\alpha-\cos\beta|&=|\cos\alpha+(-\cos\beta)|\\&\le|\cos\alpha|+|-\cos\beta|\\&=|\cos\alpha|+|\cos\beta|\\&\le2. \end{aligned}

Since $\phi^{\prime\prime}(t)$ is continuous and nonzero on $[a,b]$, then it is strictly positive or negative on $[a,b]$. Now, note that $(1/\phi^{\prime}(t))^{\prime}=-\phi^{\prime\prime}(t)/(\phi^{\prime}(t))^2$, then $(1/\phi^{\prime}(t))^{\prime}$ never changes sign on $[a,b]$. Thus, we can apply Theorem 5.5 for some $\xi\in(a,b)$: \begin{aligned} \left|\int_{a}^{b}\sin\phi(t)\,\mathrm{d}t\right|&=\left|\int_{a}^{b}\frac{1}{\phi^{\prime}(t)}\sin(\phi(t))\,\phi^{\prime}(t)\,\mathrm{d}t \right|\\ &=\left|\frac{1}{\phi^{\prime}(a)}\int_{a}^{\xi}\sin(\phi(t))\,\phi^{\prime}(t)\,\mathrm{d}t + \frac{1}{\phi^{\prime}(b)}\int_{\xi}^{b}\sin(\phi(t))\,\phi^{\prime}(t)\,\mathrm{d}t\right|\\ &\le\frac{|\cos\phi(a)-\cos\phi(\xi)|+|\cos\phi(\xi)-\cos\phi(b)|}{m}\\ &\le\frac{4}{m}. \end{aligned}\\ \hspace{14.5cm}\blacksquare $\mbox{}\\ \mbox{}\\ \scriptstyle{\text{Another way to derive the first inequality:}}$ $$\scriptstyle{ \cos\alpha - \cos\beta = 2\sin\!\left(\frac{\beta+\alpha}{2}\right)\sin\!\left(\frac{\beta-\alpha}{2}\right).}$$ $\scriptstyle{\text{Hence}}$ \begin{aligned} \scriptstyle{|\cos\alpha-\cos\beta|} &\scriptstyle{= 2 \left|\sin\!\left(\frac{\beta+\alpha}{2}\right)\right|\left|\sin\!\left(\frac{\beta-\alpha}{2}\right)\right|}\\&\scriptstyle{\le2.} \end{aligned}

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