Question

Draw a rough sketch of the curve y = $\frac{\mathrm{\pi }}{2}+2{\sin }^{2}x$ and find the area between x-axis, the curve and the ordinates x = 0, x = π.

Answer 1

x	0	$\frac{\mathrm{\pi }}{6}$	$\frac{\mathrm{\pi }}{2}$	$\frac{5\mathrm{\pi }}{6}$	$\mathrm{\pi }$
sin x	0	$\frac{1}{2}$	1	$\frac{1}{2}$	0
$y=\frac{\mathrm{\pi }}{2}+2{\sin }^{2}x$	1.57	2.07	3.57	2.07	1.57

$$\begin{gathered} y=\frac{\mathrm{\pi }}{2}+2{\sin }^{2}x\mathrm{is}\mathrm{an}\mathrm{arc}\mathrm{cutting}\mathit{}y-\mathrm{axis}\mathrm{at}(1.57,0)\mathrm{and}x=\mathrm{\pi }\mathrm{at}\left(\mathrm{\pi },1.57\right) \\ x=\mathrm{\pi }\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}y-\mathrm{axis} \\ \mathrm{Consider},\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|y\right|\mathrm{and}\mathrm{width}=dx\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{approximating}\mathrm{rectangle}=\left|y\right|dx \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=\mathrm{\pi } \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_0^{\mathrm{\pi }}\left|y\right|dx \\ \Rightarrow A={\int }_0^{\mathrm{\pi }}ydx \\ \Rightarrow A={\int }_0^{\mathrm{\pi }}\left(\frac{\mathrm{\pi }}{2}+2{\sin }^{2}x\right)dx \\ \Rightarrow A={\int }_0^{\mathrm{\pi }}\left(\frac{\mathrm{\pi }}{2}+2\left(\frac{1-\cos 2x}{2}\right)\right)dx \\ \Rightarrow A=\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}dx+{\int }_0^{\mathrm{\pi }}\left(1-\cos 2x\right)dx \\ \Rightarrow A=\frac{\mathrm{\pi }}{2}{\left[x\right]}_0^{\mathrm{\pi }}+{\left[x-\frac{\sin 2x}{2}\right]}_0^{\mathrm{\pi }} \\ \Rightarrow A=\frac{\mathrm{\pi }}{2}\left(\mathrm{\pi }\right)+\left[\mathrm{\pi }-\frac{\sin 2\mathrm{\pi }}{2}-0\right] \\ \Rightarrow A=\mathrm{\pi }\left(\frac{\mathrm{\pi }}{2}+1\right) \\ \Rightarrow A=\mathrm{\pi }\left(\frac{\mathrm{\pi }+2}{2}\right) \\ \Rightarrow A=\frac{\mathrm{\pi }}{2}\left(\mathrm{\pi }+2\right)\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{curve}\mathrm{bound}\mathrm{by}x=0\mathrm{and}x=\mathrm{\pi }\mathrm{is}\frac{\mathrm{\pi }}{2}\left(\mathrm{\pi }+2\right)\mathrm{sq}.\mathrm{units} \end{gathered}$$