Let [t] denote the greatest integer less than or equal to t . Then the value of the integral \int_-3^101([\sin (\pi x)]+e^[\cos (2 \pi x)]) d x is equal to

Definite Integration PYQ — JEE Main 2022 · 25 Jul · Shift 2 · Q29 with Solution

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Correct answer: B

We need to evaluate $I=\int_{-3}^{101}\left([\sin(\pi x)]+e^{[\cos(2\pi x)]}\right)\,dx.$

We split it as $I=I_1+I_2,$ where

\qquad I_2=\int_{-3}^{101}e^{[\cos(2\pi x)]}\,dx.$$ --- 2. Evaluate $[\sin(\pi x)]$. Since $-1\le \sin(\pi x)\le 1$: - if $\sin(\pi x)\in [0,1)$, then $[\sin(\pi x)]=0$, - if $\sin(\pi x)\in (-1,0)$, then $[\sin(\pi x)]=-1$, - at isolated points where $\sin(\pi x)=1$, value is $1$, - at isolated points where $\sin(\pi x)=0$, value is $0$, - at isolated points where $\sin(\pi x)=-1$, value is $-1$. The isolated points do not affect the integral. So effectively, $$[\sin(\pi x)]=\begin{cases} 0,& \sin(\pi x)\ge 0,\\ -1,& \sin(\pi x)<0. \end{cases}$$ Now on each interval $[n,n+1]$: - for $x\in(n,n+1)$, $\sin(\pi x)$ is positive on one half and negative on the other half, - specifically, over each unit interval, it is negative on exactly half the length. Hence over each interval of length $1$, $$\int_n^{n+1}[\sin(\pi x)]\,dx = (-1)\cdot \frac12 + 0\cdot \frac12 = -\frac12.$$ The total interval length is $$101-(-3)=104,$$ so there are $104$ unit intervals. Therefore, $$I_1=104\left(-\frac12\right)=-52.$$ --- 3. Evaluate $e^{[\cos(2\pi x)]}$. Since $-1\le \cos(2\pi x)\le 1$: - if $\cos(2\pi x)=1$ (isolated points), then $[\cos(2\pi x)]=1$, - if $0\le \cos(2\pi x)<1$, then $[\cos(2\pi x)]=0$, - if $-1\le \cos(2\pi x)<0$, then $[\cos(2\pi x)]=-1$. Again isolated points do not affect the integral, so effectively, $$e^{[\cos(2\pi x)]}=\begin{cases} 1,& \cos(2\pi x)\ge 0,\\ e^{-1}=\frac1e,& \cos(2\pi x)<0. \end{cases}$$ Over each period of length $1$, the function $\cos(2\pi x)$ is: - nonnegative for half the interval, - negative for the other half. Thus over each unit interval, $$\int_n^{n+1} e^{[\cos(2\pi x)]}\,dx =1\cdot \frac12 + \frac1e\cdot \frac12 =\frac12\left(1+\frac1e\right).$$ Since there are $104$ such intervals, $$I_2=104\cdot \frac12\left(1+\frac1e\right) =52\left(1+\frac1e\right).

Add the two parts: $I=I_1+I_2=-52+52\left(1+\frac1e\right).$

Simplifying, $I=-52+52+\frac{52}{e}=\frac{52}{e}.$

Compare with options.

$I=\frac{52}{e},$ which matches Option B.

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View written solutionFree

Correct answer: B

We need to evaluate $I=\int_{-3}^{101}\left([\sin(\pi x)]+e^{[\cos(2\pi x)]}\right)\,dx.$

We split it as $I=I_1+I_2,$ where

\qquad I_2=\int_{-3}^{101}e^{[\cos(2\pi x)]}\,dx.$$ --- 2. Evaluate $[\sin(\pi x)]$. Since $-1\le \sin(\pi x)\le 1$: - if $\sin(\pi x)\in [0,1)$, then $[\sin(\pi x)]=0$, - if $\sin(\pi x)\in (-1,0)$, then $[\sin(\pi x)]=-1$, - at isolated points where $\sin(\pi x)=1$, value is $1$, - at isolated points where $\sin(\pi x)=0$, value is $0$, - at isolated points where $\sin(\pi x)=-1$, value is $-1$. The isolated points do not affect the integral. So effectively, $$[\sin(\pi x)]=\begin{cases} 0,& \sin(\pi x)\ge 0,\\ -1,& \sin(\pi x)<0. \end{cases}$$ Now on each interval $[n,n+1]$: - for $x\in(n,n+1)$, $\sin(\pi x)$ is positive on one half and negative on the other half, - specifically, over each unit interval, it is negative on exactly half the length. Hence over each interval of length $1$, $$\int_n^{n+1}[\sin(\pi x)]\,dx = (-1)\cdot \frac12 + 0\cdot \frac12 = -\frac12.$$ The total interval length is $$101-(-3)=104,$$ so there are $104$ unit intervals. Therefore, $$I_1=104\left(-\frac12\right)=-52.$$ --- 3. Evaluate $e^{[\cos(2\pi x)]}$. Since $-1\le \cos(2\pi x)\le 1$: - if $\cos(2\pi x)=1$ (isolated points), then $[\cos(2\pi x)]=1$, - if $0\le \cos(2\pi x)<1$, then $[\cos(2\pi x)]=0$, - if $-1\le \cos(2\pi x)<0$, then $[\cos(2\pi x)]=-1$. Again isolated points do not affect the integral, so effectively, $$e^{[\cos(2\pi x)]}=\begin{cases} 1,& \cos(2\pi x)\ge 0,\\ e^{-1}=\frac1e,& \cos(2\pi x)<0. \end{cases}$$ Over each period of length $1$, the function $\cos(2\pi x)$ is: - nonnegative for half the interval, - negative for the other half. Thus over each unit interval, $$\int_n^{n+1} e^{[\cos(2\pi x)]}\,dx =1\cdot \frac12 + \frac1e\cdot \frac12 =\frac12\left(1+\frac1e\right).$$ Since there are $104$ such intervals, $$I_2=104\cdot \frac12\left(1+\frac1e\right) =52\left(1+\frac1e\right).

Add the two parts: $I=I_1+I_2=-52+52\left(1+\frac1e\right).$

Simplifying, $I=-52+52+\frac{52}{e}=\frac{52}{e}.$

Compare with options.

$I=\frac{52}{e},$ which matches Option B.

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