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Definite Integration question

2021 · 26 Feb · Shift 1 · Q29
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  5. /2021 · 26 Feb · Shift 1 · Q29

Definite Integration question

2021 · 26 Feb · Shift 1 · Q29

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∑n=1100∫n−1nex−[x]dx\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} }n=1∑100​n−1∫n​ex−[x]dx, where [ x ] is the greatest integer ≤\le≤ x, is :
  1. A
    100e
  2. B
    100(e −-− 1)
  3. C
    100(1 + e)
  4. D
    100(1 −-− e)
View written solutionFree

Correct answer: B

  1. We need to evaluate
∑n=1100∫n−1nex−[x] dx.\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]}\,dx.n=1∑100​∫n−1n​ex−[x]dx.
  1. Observe what happens on each interval [n−1,n)[n-1,n)[n−1,n):
  • For any x∈[n−1,n)x \in [n-1,n)x∈[n−1,n), we have [x]=n−1.[x]=n-1.[x]=n−1.
  • Hence x−[x]=x−(n−1).x-[x]=x-(n-1).x−[x]=x−(n−1). So on this interval,
ex−[x]=ex−(n−1).e^{x-[x]}=e^{x-(n-1)}.ex−[x]=ex−(n−1).
  1. Therefore,
∫n−1nex−[x] dx=∫n−1nex−(n−1) dx.\int_{n-1}^{n} e^{x-[x]}\,dx =\int_{n-1}^{n} e^{x-(n-1)}\,dx.∫n−1n​ex−[x]dx=∫n−1n​ex−(n−1)dx.

Let u=x−(n−1).u=x-(n-1).u=x−(n−1). Then when x=n−1x=n-1x=n−1, u=0u=0u=0, and when x=nx=nx=n, u=1u=1u=1. Thus

∫n−1nex−(n−1) dx=∫01eu du=e−1.\int_{n-1}^{n} e^{x-(n-1)}\,dx =\int_0^1 e^u\,du =e-1.∫n−1n​ex−(n−1)dx=∫01​eudu=e−1.
  1. This value is the same for every n=1,2,…,100n=1,2,\dots,100n=1,2,…,100. Hence
∑n=1100∫n−1nex−[x] dx=100(e−1).\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]}\,dx =100(e-1).n=1∑100​∫n−1n​ex−[x]dx=100(e−1).
  1. Compare with the options:
  • A: 100e100e100e
  • B: 100(e−1)100(e-1)100(e−1)
  • C: 100(1+e)100(1+e)100(1+e)
  • D: 100(1−e)100(1-e)100(1−e)

So the correct option is B.\boxed{\text{B}}.B​.

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