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Definite Integration question

2021 · 17 Mar · Shift 2 · Q28
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Definite Integration question

2021 · 17 Mar · Shift 2 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If the integral ∫010[sin⁡2πx]ex−[x]dx=αe−1+βe−12+γ\int_0^{10} {{{[\sin 2\pi x]} \over {{e^{x - [x]}}}}} dx = \alpha {e^{ - 1}} + \beta {e^{ - {1 \over 2}}} + \gamma∫010​ex−[x][sin2πx]​dx=αe−1+βe−21​+γ, where α\alphaα, β\betaβ, γ\gammaγ are integers and [x] denotes the greatest integer less than or equal to x, then the value of α\alphaα+β\betaβ+γ\gammaγ is equal to :
  1. A
    0
  2. B
    10
  3. C
    20
  4. D
    25
View written solutionFree

Correct answer: QUESTION LIKELY HAS A TYPO/INCONSISTENCY., CORRECT EVALUATED VALUE OF THE INTEGRAL IS $\DISPLAYSTYLE \FRAC{20\PI(1-E^{-1})}{1+4\PI^2}$, SO THE STORED ANSWER $0$ IS NOT JUSTIFIED FROM THE GIVEN STATEMENT.

  1. Interpret the integrand

We need to evaluate

I=∫010sin⁡(2πx)ex−[x] dx=∫010sin⁡(2πx)e−(x−[x]) dx.I=\int_0^{10} \frac{\sin(2\pi x)}{e^{x-[x]}}\,dx =\int_0^{10} \sin(2\pi x)e^{-(x-[x])}\,dx.I=∫010​ex−[x]sin(2πx)​dx=∫010​sin(2πx)e−(x−[x])dx.

Here, [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−[x]=x−n.[x]=n \quad \Rightarrow \quad x-[x]=x-n.[x]=n⇒x−[x]=x−n.

So split the integral over integer intervals.

  1. Break into sums over unit intervals

For n=0,1,2,…,9n=0,1,2,\dots,9n=0,1,2,…,9,

I=∑n=09∫nn+1sin⁡(2πx)e−(x−n)dx.I=\sum_{n=0}^{9}\int_n^{n+1} \sin(2\pi x)e^{-(x-n)}dx.I=n=0∑9​∫nn+1​sin(2πx)e−(x−n)dx.

Now substitute

t=x−n⇒x=n+t,  dx=dt,t=x-n \quad \Rightarrow \quad x=n+t, \; dx=dt,t=x−n⇒x=n+t,dx=dt,

with t∈[0,1]t\in[0,1]t∈[0,1]. Then

sin⁡(2πx)=sin⁡(2π(n+t))=sin⁡(2πn+2πt)=sin⁡(2πt),\sin(2\pi x)=\sin(2\pi(n+t))=\sin(2\pi n+2\pi t)=\sin(2\pi t),sin(2πx)=sin(2π(n+t))=sin(2πn+2πt)=sin(2πt),

since sin⁡(2πn+θ)=sin⁡θ\sin(2\pi n+\theta)=\sin\thetasin(2πn+θ)=sinθ for integer nnn.

Thus each interval gives the same value:

∫nn+1sin⁡(2πx)e−(x−n)dx=∫01e−tsin⁡(2πt)dt.\int_n^{n+1} \sin(2\pi x)e^{-(x-n)}dx =\int_0^1 e^{-t}\sin(2\pi t)dt.∫nn+1​sin(2πx)e−(x−n)dx=∫01​e−tsin(2πt)dt.

Hence

I=10∫01e−tsin⁡(2πt)dt.I=10\int_0^1 e^{-t}\sin(2\pi t)dt.I=10∫01​e−tsin(2πt)dt.
  1. Evaluate the basic integral

Let

J=∫01e−tsin⁡(2πt)dt.J=\int_0^1 e^{-t}\sin(2\pi t)dt.J=∫01​e−tsin(2πt)dt.

Use the standard result

∫eatsin⁡(bt)dt=eata2+b2(asin⁡bt−bcos⁡bt).\int e^{at}\sin(bt)dt=\frac{e^{at}}{a^2+b^2}\big(a\sin bt-b\cos bt\big).∫eatsin(bt)dt=a2+b2eat​(asinbt−bcosbt).

Here a=−1a=-1a=−1, b=2πb=2\pib=2π. Therefore

J=[e−t1+4π2(−sin⁡(2πt)−2πcos⁡(2πt))]01.J=\left[\frac{e^{-t}}{1+4\pi^2}\big(-\sin(2\pi t)-2\pi\cos(2\pi t)\big)\right]_0^1.J=[1+4π2e−t​(−sin(2πt)−2πcos(2πt))]01​.

Now evaluate at the limits:

  • At t=1t=1t=1:
sin(2π)=0,cos(2π)=1,sin(2\pi)=0, \qquad cos(2\pi)=1,sin(2π)=0,cos(2π)=1,

so

e−11+4π2(0−2π)=−2πe−11+4π2.\frac{e^{-1}}{1+4\pi^2}(0-2\pi)= -\frac{2\pi e^{-1}}{1+4\pi^2}.1+4π2e−1​(0−2π)=−1+4π22πe−1​.
  • At t=0t=0t=0:
sin0=0,cos0=1,sin 0=0, \qquad cos 0=1,sin0=0,cos0=1,

so

11+4π2(0−2π)=−2π1+4π2.\frac{1}{1+4\pi^2}(0-2\pi)= -\frac{2\pi}{1+4\pi^2}.1+4π21​(0−2π)=−1+4π22π​.

Thus

J=−2πe−11+4π2+2π1+4π2=2π(1−e−1)1+4π2.J=-\frac{2\pi e^{-1}}{1+4\pi^2}+\frac{2\pi}{1+4\pi^2} =\frac{2\pi(1-e^{-1})}{1+4\pi^2}.J=−1+4π22πe−1​+1+4π22π​=1+4π22π(1−e−1)​.

Therefore

I=10J=20π(1−e−1)1+4π2.I=10J=\frac{20\pi(1-e^{-1})}{1+4\pi^2}.I=10J=1+4π220π(1−e−1)​.
  1. Compare with the given form

The question claims

I=αe−1+βe−1/2+γ,I=\alpha e^{-1}+\beta e^{-1/2}+\gamma,I=αe−1+βe−1/2+γ,

where α,β,γ\alpha,\beta,\gammaα,β,γ are integers.

But our exact value is

I=20π1+4π2−20π1+4π2e−1,I=\frac{20\pi}{1+4\pi^2}-\frac{20\pi}{1+4\pi^2}e^{-1},I=1+4π220π​−1+4π220π​e−1,

which involves the non-integer coefficient 20π1+4π2\dfrac{20\pi}{1+4\pi^2}1+4π220π​ and has no e−1/2e^{-1/2}e−1/2 term.

So the expression stated in the problem is inconsistent with the actual integral.

If we still compare numerically with the options, we can test whether the integral could correspond to any of them.

Since

I≈20π(1−1/e)1+4π2≈0.983,I\approx \frac{20\pi(1-1/e)}{1+4\pi^2}\approx 0.983,I≈1+4π220π(1−1/e)​≈0.983,

it cannot determine an integer sum like 0,10,20,250,10,20,250,10,20,25 from valid integer α,β,γ\alpha,\beta,\gammaα,β,γ in the stated form.

This strongly suggests a typo in the question statement. However, among the provided options, the stored answer is A: 0.

  1. Conclusion

From direct evaluation, the integral equals

20π(1−e−1)1+4π2.\boxed{\frac{20\pi(1-e^{-1})}{1+4\pi^2}}.1+4π220π(1−e−1)​​.

So the given representation

αe−1+βe−1/2+γ\alpha e^{-1}+\beta e^{-1/2}+\gammaαe−1+βe−1/2+γ

with integer α,β,γ\alpha,\beta,\gammaα,β,γ is not valid as written.

Hence I do not agree with the stored answer.

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