area bounded by y=[sin x+cos x] and the lines x=0,x=pi is (where [.] denotes the greatest integer function)

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{y}=\left[\mathrm{sinx}+\mathrm{cosx}\right] \\ \mathrm{Required} \mathrm{area} \mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}: \\ \mathrm{A}={\int }_0^{\mathrm{\pi }}\mathrm{y}.\mathrm{dx} \\ ={\int }_0^{\mathrm{\pi }}\left[\mathrm{sinx}+\mathrm{cosx}\right].\mathrm{dx} \\ ={\int }_0^{\mathrm{\pi }}\left[\sqrt{2}\left(\mathrm{sinx}.\frac{1}{\sqrt{2}}+\mathrm{cosx}.\frac{1}{\sqrt{2}}\right)\right].\mathrm{dx} \\ ={\int }_0^{\mathrm{\pi }}\left[\sqrt{2}\left(\mathrm{sinx}.\cos \left(\frac{\mathrm{\pi }}{4}\right)+\mathrm{cosx}.\sin \left(\frac{\mathrm{\pi }}{4}\right)\right)\right].\mathrm{dx} \\ ={\int }_0^{\mathrm{\pi }}\left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right].\mathrm{dx} \left[\mathrm{sinA}.\mathrm{cosB}+\mathrm{cosA}.\mathrm{sinB}=\sin \left(\mathrm{A}+\mathrm{B}\right)\right] \\ \mathrm{As} 0<\mathrm{x}<\mathrm{\pi } \\ \Rightarrow 0+\frac{\mathrm{\pi }}{4}<\mathrm{x}+\frac{\mathrm{\pi }}{4}<\mathrm{\pi }+\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{\mathrm{\pi }}{4}<\mathrm{x}+\frac{\mathrm{\pi }}{4}<\frac{5\mathrm{\pi }}{4} \\ \mathrm{When} \\ 0<\mathrm{x}\le \frac{\mathrm{\pi }}{2} \\ \frac{\mathrm{\pi }}{4}<\mathrm{x}+\frac{\mathrm{\pi }}{4}\le \frac{3\mathrm{\pi }}{4} \\ \mathrm{Hence} \sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\in [1,\sqrt{2}) \\ \Rightarrow \left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right]=1 \\ \mathrm{When} \\ \frac{\mathrm{\pi }}{2}<\mathrm{x}\le \frac{3\mathrm{\pi }}{4} \\ \frac{3\mathrm{\pi }}{4}<\mathrm{x}+\frac{\mathrm{\pi }}{4}\le \mathrm{\pi } \\ \mathrm{Hence} \sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\in [0,1) \\ \Rightarrow \left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right]=0 \\ \mathrm{When} \\ \frac{3\mathrm{\pi }}{4}<\mathrm{x}<\mathrm{\pi } \\ \mathrm{\pi }<\mathrm{x}+\frac{\mathrm{\pi }}{4}<\frac{5\mathrm{\pi }}{4} \\ \mathrm{Hence} \sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\in \left(-1,0\right) \\ \Rightarrow \left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right]=-1 \\ \mathrm{Now} \\ {\int }_0^{\mathrm{\pi }}\left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right].\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right].\mathrm{dx}+{\int }_{\frac{\mathrm{\pi }}{2}}^{\frac{3\mathrm{\pi }}{4}}\left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right].\mathrm{dx}+{\int }_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }}\left[\sqrt{2}\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)\right].\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1.\mathrm{dx}+{\int }_{\frac{\mathrm{\pi }}{2}}^{\frac{3\mathrm{\pi }}{4}}0.\mathrm{dx}+{\int }_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }}\left(-1\right).\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1.\mathrm{dx}-{\int }_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }}1.\mathrm{dx} \\ ={\left[\mathrm{x}\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[\mathrm{x}\right]}_{\frac{3\mathrm{\pi }}{4}}^{\mathrm{\pi }} \\ =\left[\frac{\mathrm{\pi }}{2}-0\right]-\left[\mathrm{\pi }-\frac{3\mathrm{\pi }}{4}\right] \\ =\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{4} \\ =\frac{\mathrm{\pi }}{4} \\ {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\pi }}\left[\mathbf{sinx}\mathbf{+}\mathbf{cosx}\right]\mathbf{.}\mathbf{dx}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{4}} \end{gathered}$$

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