Let the area of the region enclosed by the curve y=\min sin x, \cos x\ and the x axis between x=-\pi to x=\pi be A . Then A^2 is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q59 with Solution

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Correct answer: 16

We need the area enclosed by $y=\min\{\sin x,\cos x\}$ and the $x$ -axis for $x\in[-\pi,\pi]$ .
First, find where $\sin x=\cos x$ : $\tan x=1 \implies x=\frac{\pi}{4}+n\pi.$ In $[-\pi,\pi]$ , the relevant points are $x=-\frac{3\pi}{4},\quad x=\frac{\pi}{4}.$
Determine which of $\sin x,\cos x$ is smaller on each interval.
- For $x\in[-\pi,-3\pi/4]$ , take $x=-\pi$ : $\sin(-\pi)=0,\quad \cos(-\pi)=-1,$ so $\min\{\sin x,\cos x\}=\cos x$ .
- For $x\in[-3\pi/4,\pi/4]$ , take $x=0$ : $\sin 0=0,\quad \cos 0=1,$ so $\min\{\sin x,\cos x\}=\sin x$ .
- For $x\in[\pi/4,\pi]$ , take $x=\pi/2$ : $\sin\frac{\pi}{2}=1,\quad \cos\frac{\pi}{2}=0,$ so $\min\{\sin x,\cos x\}=\cos x$ .
Hence,
\begin{cases} \cos x, & -\pi\le x\le -\frac{3\pi}{4},\\[4pt] \sin x, & -\frac{3\pi}{4}\le x\le \frac{\pi}{4},\\[4pt] \cos x, & \frac{\pi}{4}\le x\le \pi. \end{cases}$$
Since area with the $x$ -axis means geometric area, we compute $A=\int_{-\pi}^{\pi}\left|\min\{\sin x,\cos x\}\right|dx.$

Break it into intervals where sign is easy:

$A=\int_{-\pi}^{-3\pi/4}|\cos x|dx+\int_{-3\pi/4}^{\pi/4}|\sin x|dx+\int_{\pi/4}^{\pi}|\cos x|dx.$
Evaluate each part.

(i) On $[-\pi,-3\pi/4]$ , $\cos x<0$ , so $|\cos x|=-\cos x$ :
$=\int_{-\pi}^{-3\pi/4}(-\cos x)dx =\left[-\sin x\right]_{-\pi}^{-3\pi/4}$$ $$=-\sin\left(-\frac{3\pi}{4}\right)+\sin(-\pi) =-\left(-\frac{\sqrt2}{2}\right)+0 =\frac{\sqrt2}{2}.$$ **(ii)** On $[-3\pi/4,0]$, $\sin x<0$, and on $[0,\pi/4]$, $\sin x>0$: $$\int_{-3\pi/4}^{\pi/4}|\sin x|dx =\int_{-3\pi/4}^{0}(-\sin x)dx+\int_{0}^{\pi/4}\sin x\,dx$$ $$=\left[\cos x\right]_{-3\pi/4}^{0}+\left[-\cos x\right]_{0}^{\pi/4}$$ $$=(1-(-\tfrac{\sqrt2}{2})) + \left(-\tfrac{\sqrt2}{2}+1\right) =\left(1+\frac{\sqrt2}{2}\right)+\left(1-\frac{\sqrt2}{2}\right)=2.$$ **(iii)** On $[\pi/4,\pi/2]$, $\cos x>0$, and on $[\pi/2,\pi]$, $\cos x<0$: $$\int_{\pi/4}^{\pi}|\cos x|dx =\int_{\pi/4}^{\pi/2}\cos x\,dx+\int_{\pi/2}^{\pi}(-\cos x)dx$$ $$=\left[\sin x\right]_{\pi/4}^{\pi/2}+\left[-\sin x\right]_{\pi/2}^{\pi}$$ $$=(1-\tfrac{\sqrt2}{2})+(0-(-1)) =1-\frac{\sqrt2}{2}+1 =2-\frac{\sqrt2}{2}.$$$
Add all areas: $A=\frac{\sqrt2}{2}+2+\left(2-\frac{\sqrt2}{2}\right)=4.$
Therefore, $A^2=4^2=16.$
Comparison with stored answer:
- Derived answer: $16$
- Stored correct answer: $16$
- They match.

More from Area Under the Curves

View written solutionFree

Correct answer: 16

We need the area enclosed by $y=\min\{\sin x,\cos x\}$ and the $x$ -axis for $x\in[-\pi,\pi]$ .
First, find where $\sin x=\cos x$ : $\tan x=1 \implies x=\frac{\pi}{4}+n\pi.$ In $[-\pi,\pi]$ , the relevant points are $x=-\frac{3\pi}{4},\quad x=\frac{\pi}{4}.$
Determine which of $\sin x,\cos x$ is smaller on each interval.
- For $x\in[-\pi,-3\pi/4]$ , take $x=-\pi$ : $\sin(-\pi)=0,\quad \cos(-\pi)=-1,$ so $\min\{\sin x,\cos x\}=\cos x$ .
- For $x\in[-3\pi/4,\pi/4]$ , take $x=0$ : $\sin 0=0,\quad \cos 0=1,$ so $\min\{\sin x,\cos x\}=\sin x$ .
- For $x\in[\pi/4,\pi]$ , take $x=\pi/2$ : $\sin\frac{\pi}{2}=1,\quad \cos\frac{\pi}{2}=0,$ so $\min\{\sin x,\cos x\}=\cos x$ .
Hence,
\begin{cases} \cos x, & -\pi\le x\le -\frac{3\pi}{4},\\[4pt] \sin x, & -\frac{3\pi}{4}\le x\le \frac{\pi}{4},\\[4pt] \cos x, & \frac{\pi}{4}\le x\le \pi. \end{cases}$$
Since area with the $x$ -axis means geometric area, we compute $A=\int_{-\pi}^{\pi}\left|\min\{\sin x,\cos x\}\right|dx.$

Break it into intervals where sign is easy:

$A=\int_{-\pi}^{-3\pi/4}|\cos x|dx+\int_{-3\pi/4}^{\pi/4}|\sin x|dx+\int_{\pi/4}^{\pi}|\cos x|dx.$
Evaluate each part.

(i) On $[-\pi,-3\pi/4]$ , $\cos x<0$ , so $|\cos x|=-\cos x$ :
$=\int_{-\pi}^{-3\pi/4}(-\cos x)dx =\left[-\sin x\right]_{-\pi}^{-3\pi/4}$$ $$=-\sin\left(-\frac{3\pi}{4}\right)+\sin(-\pi) =-\left(-\frac{\sqrt2}{2}\right)+0 =\frac{\sqrt2}{2}.$$ **(ii)** On $[-3\pi/4,0]$, $\sin x<0$, and on $[0,\pi/4]$, $\sin x>0$: $$\int_{-3\pi/4}^{\pi/4}|\sin x|dx =\int_{-3\pi/4}^{0}(-\sin x)dx+\int_{0}^{\pi/4}\sin x\,dx$$ $$=\left[\cos x\right]_{-3\pi/4}^{0}+\left[-\cos x\right]_{0}^{\pi/4}$$ $$=(1-(-\tfrac{\sqrt2}{2})) + \left(-\tfrac{\sqrt2}{2}+1\right) =\left(1+\frac{\sqrt2}{2}\right)+\left(1-\frac{\sqrt2}{2}\right)=2.$$ **(iii)** On $[\pi/4,\pi/2]$, $\cos x>0$, and on $[\pi/2,\pi]$, $\cos x<0$: $$\int_{\pi/4}^{\pi}|\cos x|dx =\int_{\pi/4}^{\pi/2}\cos x\,dx+\int_{\pi/2}^{\pi}(-\cos x)dx$$ $$=\left[\sin x\right]_{\pi/4}^{\pi/2}+\left[-\sin x\right]_{\pi/2}^{\pi}$$ $$=(1-\tfrac{\sqrt2}{2})+(0-(-1)) =1-\frac{\sqrt2}{2}+1 =2-\frac{\sqrt2}{2}.$$$
Add all areas: $A=\frac{\sqrt2}{2}+2+\left(2-\frac{\sqrt2}{2}\right)=4.$
Therefore, $A^2=4^2=16.$
Comparison with stored answer:
- Derived answer: $16$
- Stored correct answer: $16$
- They match.

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