how do i tackle such problems?

Let f(x) be a function differentiable on the interval$ $$\Bigl[$$ 0, \frac\pi2 $$\Bigr]$$ $ such that f'(x) is integrable on this interval.

Prove the following assertions:


If $$\int^{\frac{\pi}{2}}_{0} f(x) \cos x dx \lt f \left(\frac\pi2\right),$$ then $$ \int^{\frac{\pi}{2}}_{0} f'(x) \sin x dx \gt 0. $$


If $$f\left(\frac\pi3\right) = \sqrt{3}f\left(\frac\pi6\right),$$ then $$ \int^{\frac{\pi}{3}}_{\frac{\pi}{6}} f(x) sinx dx = \int^{\frac{\pi}{3}}_{\frac{\pi}{6}} f'(x) cosx dx.$$

  • $\begingroup$ Hint: Use integration by parts for both. $\endgroup$ – Ivan Loh Apr 20 '14 at 16:21

For (1), using integration by parts we have \begin{align} \int_{0}^{\pi/2}{f(x)\cos x dx} \ & = \left. f(x)\sin x \right|_{0}^{\pi/2}-\int_{0}^{\pi/2}{f'(x)\sin x dx} \\ & =f(\pi/2)\sin(\pi/2)-0-\int_{0}^{\pi/2}{f'(x)\sin x dx} \\ & =f(\pi/2)-\int_{0}^{\pi/2}{f'(x)\sin x dx} \end{align} Then $\displaystyle{\int_{0}^{\pi/2}{f(x)\cos x dx}}<f(\pi/2)$ implies $\displaystyle{\int_{0}^{\pi/2}{f'(x)\sin x dx}}>0$.

For (2) we can use integration by parts too. \begin{align} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f(x)\sin x dx} & = \left.-f(x)\cos x\right|_{\frac{\pi}{6}}^{\frac{\pi}{3}} + \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f'(x)\cos x dx} \\ & = -f(\pi/3)\cos(\pi/3)-[-f(\pi/6)\cos (\pi/6)] + \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f'(x)\cos x dx}\\ & = -\frac{1}{2}f(\pi/3)+\frac{\sqrt{3}}{2}f(\pi/6)+ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f'(x)\cos x dx}\\ \end{align} From here it is clear that $f(\frac{\pi}{3})=\sqrt{3}f(\frac{\pi}{6})$ implies $\displaystyle{\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f(x)\sin x dx}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}{f'(x)\cos x dx}}$.

  • $\begingroup$ what about (2)? $\endgroup$ – adsisco Apr 21 '14 at 6:41

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