Question

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

Answer 1

X	0	`pi/6`	`pi/2`	`(5pi)/6`	`pi`
sin x	0	`1/2`	1	`1/2`	0
\[y = \frac{\pi}{2} + 2 \sin^2 x\]	1.57	2.07	3.57	2.07	1.57

\[y = \frac{\pi}{2} + 2 \sin^2 x \text{ is an arc cutting }y - \text{ axis at }(1 . 57, 0 ) \text{ and }x = \pi at \left( \pi, 1 . 57 \right)\]
\[x = \pi \text{ is a line parallel to }y - \text{ axis }\]
\[ \text{ Consider, a vertical strip of length }= \left| y \right| \text{ and width }= dx \text{ in the first quadrant }\]
\[ \therefore \text{ Area of the approximating rectangle }= \left| y \right| dx \]
\[ \text{ The approximating rectangle moves from }x = 0 \text{ to }x = \pi \]
\[ \Rightarrow \text{ Area of the shaded region }= \int_0^\pi \left| y \right| dx \]
\[ \Rightarrow A = \int_0^\pi y dx \]
\[ \Rightarrow A = \int_0^\pi \left( \frac{\pi}{2} + 2 \sin^2 x \right) dx\]
\[ \Rightarrow A = \int_0^\pi \left( \frac{\pi}{2} + 2\left( \frac{1 - \cos 2x}{2} \right) \right) dx\]
\[ \Rightarrow A = \frac{\pi}{2} \int_0^\pi dx + \int_0^\pi \left( 1 - \cos 2x \right) dx\]
\[ \Rightarrow A = \frac{\pi}{2} \left[ x \right]_0^\pi + \left[ x - \frac{\sin 2x}{2} \right]_0^\pi \]
\[ \Rightarrow A = \frac{\pi}{2}\left( \pi \right) + \left[ \pi - \frac{\sin 2\pi}{2} - 0 \right]\]
\[ \Rightarrow A = \pi \left( \frac{\pi}{2} + 1 \right)\]
\[ \Rightarrow A = \pi \left( \frac{\pi + 2}{2} \right)\]
\[ \Rightarrow A = \frac{\pi}{2} \left( \pi + 2 \right) \text{ sq . units }\]
\[ \therefore \text{ Area of curve bound by }x = 0 \text{ and }x = \pi \text{ is }\frac{\pi}{2} \left( \pi + 2 \right) \text{ sq . units }\]