Question

Compare the areas under the curves y = cos² x and y = sin² x between x = 0 and x = π.

Answer 1

X	0	\[\frac{\pi}{4}\]	\[\frac{\pi}{3}\]	\[\frac{\pi}{2}\]	\[\frac{2\pi}{3}\]	\[\frac{5\pi}{6}\]	\[\pi\]
\[y = \cos^2 x\]	1	0.5	0.25	0	0.25	0.75	1
\[y = \sin^2 x\]	0	0.5	0.75	1	0.75	0.25	0

Let A₁ be the area of curve \[y = \cos^2 x\text{ between }x = 0 \text{ and }x = \pi\]

Let A₂ be the area of curve \[y = \sin^2 x \text{ between }x = 0\text{ and }x = \pi\]

Consider, a vertical strip of length \[= \left| y \right|\] and width \[= dx\]  in the shaded region of both the curves

The area of approximating rectangle \[= \left| y \right| dx\]

\[\text{The approximating rectangle moves from}x = 0\text{ to }x = \pi\]
\[ A_1 = \int_0^\pi \left| y \right| dx\]
\[ \Rightarrow A_1 = \int_0^\pi y dx ..................\left[ 0 \leq x \leq \pi , y > 0 \Rightarrow \left| y \right| = y \right]\]
\[ \Rightarrow A_1 = \int_0^\pi \cos^2 x dx\]
\[ \Rightarrow A_1 = \int_0^\pi \left( 1 + cos 2x \right) dx .................\left[ \cos^2 x = \left( 1 + \cos 2x \right) \right]\]
\[ \Rightarrow A_1 = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_0^\pi \]
\[ \Rightarrow A_1 = \frac{1}{2}\left[ \pi + \frac{\sin 2\pi}{2} - 0 \right]\]
\[ \Rightarrow A_1 = \frac{\pi}{2} \text{ Sq . units }\]
Also, 
\[ A_2 = \int_0^\pi \left| y \right| dx\]
\[ \Rightarrow A_2 = \int_0^\pi y dx .................\left[ 0 \leq x \leq \pi , y > 0 \Rightarrow \left| y \right| = y \right]\]
\[ \Rightarrow A_2 = \int_0^\pi \sin^2 x dx\]
\[ \Rightarrow A_2 = \left[ \frac{x}{2} - \frac{1}{2}\frac{\sin 2x}{2} \right]_0^\pi \]
\[ \Rightarrow A_2 = \frac{\pi}{2} - \left( \frac{1}{2}\frac{\sin 2\pi}{2} \right)\]
\[ \Rightarrow A_2 = \frac{\pi}{2} sq . units\]
\[ \therefore\text{ Area of curves }y = \cos^2 x\text{ and area of curve }y = \sin^2 x \text{ are both equal to }\frac{\pi}{2}\text{ sq . units }\]