How can we prove this integral inequality?

Question

There is the first inequality that I should solve: For all $R>0$ $$\int_0^{\pi/2}e^{-R\sin(\varphi)}d\varphi < \frac\pi{2R}(1 - e^{-R})$$

But in the next exercise, there is a more general case: if $f''(x) < 0 $ on $[a, b]$, then $$ \int_a^b e^{-\lambda f(x)} dx < \frac{b-a}{\lambda(f(b) - f(a))}(e^{-\lambda f(b)} - e^{-\lambda f(a)}) $$

First of all I think that i should apply Jensen's inequality for integrals, because $ e^{-\lambda f(x)} $ is convex function, when $ f''(x) < 0 $. But it is provide me wrong inequation ( $\int >= $ something). So I don't know what to do. Please help me solve this inequality.

Answer 1

Let $\lambda > 0$. Since $f''(x) < 0$ on $[a, b]$, it lies upper than the line segment given by equation $g(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$. Therefore, $\forall x\in[a,b]:f(x)\geqslant g(x) \implies e^{-\lambda f(x)} \leqslant e^{-\lambda g(x)}$. From monotony of the Riemann integral we obtain: \begin{equation} \int_a^b{e^{-\lambda f(x)}dx}<\int_a^b{e^{-\lambda g(x)}dx}=\frac{b-a}{\lambda{(f(b)-f(a))}}(e^{-\lambda f(a)}-e^{-\lambda f(b)}). \end{equation}

Answer 2

For first inequality consider the function g(x)=2x/pi , because sin(x) is concave in (0,pi/2) then we have that sin(x) >= g(x), thus e^(-Rsin(x)) <= e^(-2Rx/pi) and integrating you get the result.