Mathematics - Area Under Curves Question with Solution | TestHub

For a fixed $\alpha >1,{\alpha }^n\to \infty$ and ${\alpha }^{-n}\to 0$ as $n\to \infty$, so the ratio tends to $|\sin x|:a=|\sin x|$. For $\alpha <1,{\alpha }^n\to 0,{\alpha }^{-n}\to \infty$, so $b=|\cos x|.c=\frac{\pi }{4}{\int }_0^1\cos \frac{\pi t}{2}dt=\frac{\pi }{4}\cdot \frac{2}{\pi }=\frac{1}{2}$. Since $\max (|\sin x|,|\cos x|)\ge$ $\frac{1}{\sqrt{2}}>\frac{1}{2},f(x)=\max \{|\sin x|,|\cos x|\}$. Area $={\int }_0^{\pi /4}\cos xdx+{\int }_{\pi /4}^{3\pi /4}\sin xdx+{\int }_{3\pi /4}^{\pi }|\cos x|dx=\frac{\sqrt{2}}{2}+$ $\sqrt{2}+\frac{\sqrt{2}}{2}=2\sqrt{2}$.