Question

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

Answer 1

Consider the value of y for different values of x

x	0	$\frac{\mathrm{\pi }}{4}$	$\frac{\mathrm{\pi }}{3}$	$\frac{\mathrm{\pi }}{2}$	$\frac{2\mathrm{\pi }}{3}$	$\frac{5\mathrm{\pi }}{6}$	$\pi$
$y={\cos }^{2}x$	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 $y={\cos }^{2}x\mathrm{between}x=0\mathrm{and}x=\mathrm{\pi }$
Let A₂ be the area of curve $y={\sin }^{2}x\mathrm{between}x=0\mathrm{and}x=\pi$

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 $=\left|y\right|dx$

$$\begin{gathered} \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=\mathrm{\pi } \\ {A}_1={\int }_0^{\mathrm{\pi }}\left|y\right|dx \\ \Rightarrow A_1={\int }_0^{\mathrm{\pi }}ydx\left[0\le x\le \pi ,y>0\Rightarrow \left|y\right|=y\right] \\ \Rightarrow A_1={\int }_0^{\mathrm{\pi }}{\cos }^{2}xdx \\ \Rightarrow A_1={\int }_0^{\mathrm{\pi }}\left(1+cos2x\right)dx\left[{\cos }^{2}x=\left(1+\cos 2x\right)\right] \\ \Rightarrow A_1=\frac{1}{2}{\left[x+\frac{\sin 2x}{2}\right]}_0^{\mathrm{\pi }} \\ \Rightarrow A_1=\frac{1}{2}\left[\mathrm{\pi }+\frac{\sin 2\mathrm{\pi }}{2}-0\right] \\ \Rightarrow A_1=\frac{\mathrm{\pi }}{2}\mathrm{Sq}.\mathrm{units} \\ \mathrm{Also}, \\ {A}_2={\int }_0^{\mathrm{\pi }}\left|y\right|dx \\ \Rightarrow A_2={\int }_0^{\mathrm{\pi }}ydx\left[0\le x\le \pi ,y>0\Rightarrow \left|y\right|=y\right] \\ \Rightarrow A_2={\int }_0^{\mathrm{\pi }}{\sin }^{2}xdx \\ \Rightarrow A_2={\left[\frac{x}{2}-\frac{1}{2}\frac{\sin 2x}{2}\right]}_0^{\mathrm{\pi }} \\ \Rightarrow A_2=\frac{\mathrm{\pi }}{2}-\left(\frac{1}{2}\frac{\sin 2\mathrm{\pi }}{2}\right) \\ \Rightarrow A_2=\frac{\pi }{2}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{curves}y={\cos }^{2}x\mathrm{and}\mathrm{area}\mathrm{of}\mathrm{curve}y={\sin }^{2}x\mathrm{are}\mathrm{both}\mathrm{equal}\mathrm{to}\frac{\mathrm{\pi }}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$