Question
Compare the area under the curve $\text{y}=\cos^2\text{x}\text{ and }\text{y}=\sin^2\text{x}$ between $x = 0$ and $\text{x}=\pi.$
✓
Answer
Consider the value of y for different values of x
| $\text{x}$ | $0$ | $\frac{\pi}{4}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\frac{2\pi}{3}$ | $\frac{5\pi}{6}$ | $\pi$ |
| $\text{y}=\cos^2\text{x}$ | $1$ | $0.5$ | $ 0.25$ | $0$ | $ 0.25$ | $ 0.75$ | $1$ |
| $$$\text{y}=\sin^2\text{x}$ | $0$ | $0.5$ | $ 0.75$ | $1$ | $ 0.75$ | $0.25$ | $0$ |
Let $A_2$ be the area of curve $\text{y}=\sin^2\text{x}$ between x = 0 and $\text{x}=\pi$
Consider, a vertical strip of lenght = |y| and width = $\text{dx}$ in the shaded region of both the curves The area of approximating $=|\text{y}|\text{dx}$
The approximating rectangle moves from x = 0 to $\text{x}=\pi$
$\text{A}_1=\int\limits^\pi_0\text{|y| dx}$
$[0\leq\text{x}\leq\pi,\text{ y}>0\Rightarrow|\text{y}|=\text{y}]$
$\Rightarrow\text{A}_1=\int\limits^\pi_0\cos^2\text{x dx}$
$\Rightarrow\text{A}_1=\int\limits^\pi_0\text{y dx}$
$\Rightarrow\text{A}_1=\int\limits^\pi_0(1+\cos2\text{x})\text{dx}$
$\Rightarrow\text{A}_1=\frac{1}{2}\big[\text{x}+\frac{\sin2\text{x}}{2}\big]^\pi_0$
$\Rightarrow\text{A}_1=\frac{1}{2}\big[\text{x}+\frac{\sin2\text{x}}{2}-0\big]$
$\Rightarrow\text{A}_1=\frac{\pi}{2}\text{sq. units}$
Also,
$\text{A}_2=\int\limits^\pi_0\text{|y| dx}$
$\Rightarrow\text{A}_2=\int\limits^\pi_0\text{y dx}$
$\Rightarrow\text{A}_2=\int\limits^\pi_0\sin^2\text{x dx}$
$\Rightarrow\text{A}_2=\bigg[\frac{x}{2}-\frac{1}{2}\frac{\sin2\text{}x}{2}\bigg]^\pi_0$
$\Rightarrow\text{A}_2=\frac{x}{2}-\bigg(\frac{1}{2}\frac{\sin2\pi}{2}\bigg)$
$\Rightarrow\text{A}_2=\frac{\pi}{2}\text{sq. units}$
$\therefore$ Area of curves $\text{y}=\cos^2\text{x}$ and area of curve $\text{y}=\sin^2\text{x}$ are both equal to $\frac{\pi}{2}\text{sq. units}$
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