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Definite Integration question

2021 · 26 Aug · Shift 2 · Q26
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  5. /2021 · 26 Aug · Shift 2 · Q26

Definite Integration question

2021 · 26 Aug · Shift 2 · Q26

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If the value of the integral ∫05x+[x]ex−[x]dx=αe−1+β\int\limits_0^5 {{{x + [x]} \over {{e^{x - [x]}}}}dx = \alpha {e^{ - 1}} + \beta }0∫5​ex−[x]x+[x]​dx=αe−1+β, where α\alphaα, β∈\beta\inβ∈ R, 5 α\alphaα + 6 β\betaβ = 0, and [x] denotes the greatest integer less than or equal to x; then the value of (α\alphaα+β\betaβ)2 is equal to :
  1. A
    100
  2. B
    25
  3. C
    16
  4. D
    36
View written solutionFree

Correct answer: B

  1. Given integral

We need to evaluate I=∫05x+[x]ex−[x] dx.I=\int_0^5 \frac{x+[x]}{e^{x-[x]}}\,dx.I=∫05​ex−[x]x+[x]​dx.

Since [x][x][x] is the greatest integer function, on each interval [n,n+1)[n,n+1)[n,n+1) we have [x]=n,x=n+t(0≤t<1).[x]=n, \qquad x=n+t \quad (0\le t<1).[x]=n,x=n+t(0≤t<1). Then x−[x]=t,x+[x]=(n+t)+n=2n+t.x-[x]=t, \qquad x+[x]=(n+t)+n=2n+t.x−[x]=t,x+[x]=(n+t)+n=2n+t. So the integral breaks as I=∑n=04∫nn+1x+nex−n dx.I=\sum_{n=0}^4 \int_n^{n+1} \frac{x+n}{e^{x-n}}\,dx.I=∑n=04​∫nn+1​ex−nx+n​dx. Using x=n+tx=n+tx=n+t, dx=dtdx=dtdx=dt: I=∑n=04∫01(2n+t)e−t dt.I=\sum_{n=0}^4 \int_0^1 (2n+t)e^{-t}\,dt.I=∑n=04​∫01​(2n+t)e−tdt.

  1. Separate the sum

I=∑n=04(2n∫01e−t dt+∫01te−t dt).I=\sum_{n=0}^4 \left(2n\int_0^1 e^{-t}\,dt+\int_0^1 te^{-t}\,dt\right).I=∑n=04​(2n∫01​e−tdt+∫01​te−tdt).

Now, ∫01e−t dt=[−e−t]01=1−e−1.\int_0^1 e^{-t}\,dt=\left[-e^{-t}\right]_0^1=1-e^{-1}.∫01​e−tdt=[−e−t]01​=1−e−1.

Also, ∫01te−t dt\int_0^1 te^{-t}\,dt∫01​te−tdt by integration by parts:

Take u=t,dv=e−tdt⇒du=dt,v=−e−t.u=t,\quad dv=e^{-t}dt \Rightarrow du=dt,\quad v=-e^{-t}.u=t,dv=e−tdt⇒du=dt,v=−e−t. Then ∫te−tdt=−te−t+∫e−t(−1)dt=−(t+1)e−t.\int te^{-t}dt=-te^{-t}+\int e^{-t}(-1)dt=-(t+1)e^{-t}.∫te−tdt=−te−t+∫e−t(−1)dt=−(t+1)e−t. Hence ∫01te−t dt=[−(t+1)e−t]01=1−2e=1−2e−1.\int_0^1 te^{-t}\,dt=\left[-(t+1)e^{-t}\right]_0^1=1-\frac{2}{e}=1-2e^{-1}.∫01​te−tdt=[−(t+1)e−t]01​=1−e2​=1−2e−1.

  1. Compute the sum

Since ∑n=042n=2(0+1+2+3+4)=2⋅10=20,\sum_{n=0}^4 2n=2(0+1+2+3+4)=2\cdot 10=20,∑n=04​2n=2(0+1+2+3+4)=2⋅10=20, and there are 555 identical ∫01te−tdt\int_0^1 te^{-t}dt∫01​te−tdt terms,

I=20(1−e−1)+5(1−2e−1).I=20(1-e^{-1})+5(1-2e^{-1}).I=20(1−e−1)+5(1−2e−1).

So, I=20−20e−1+5−10e−1=25−30e−1.I=20-20e^{-1}+5-10e^{-1}=25-30e^{-1}.I=20−20e−1+5−10e−1=25−30e−1.

Thus in the form I=αe−1+β,I=\alpha e^{-1}+\beta,I=αe−1+β, we get α=−30,β=25.\alpha=-30,\qquad \beta=25.α=−30,β=25.

  1. Check the given relation

5α+6β=5(−30)+6(25)=−150+150=0,5\alpha+6\beta=5(-30)+6(25)=-150+150=0,5α+6β=5(−30)+6(25)=−150+150=0, which is satisfied.

  1. Find (α+β)2(\alpha+\beta)^2(α+β)2

α+β=−30+25=−5.\alpha+\beta=-30+25=-5.α+β=−30+25=−5. Therefore, (α+β)2=(−5)2=25. (\alpha+\beta)^2 = (-5)^2=25.(α+β)2=(−5)2=25.

  1. Option check

The correct option is: 25\boxed{25}25​ which is Option B.

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