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Definite Integration question

2025 · 2 Apr · Shift 1 · Q50
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Definite Integration question

2025 · 2 Apr · Shift 1 · Q50

JEE MainMathematicsDefinite IntegrationNumerical+4 / −1
Let [.] denote the greatest integer function. If ∫0e3[1ex−1]dx=α−log⁡e2\int\limits_0^{e^3}\left[\frac{1}{e^{x-1}}\right] d x=\alpha-\log _e 20∫e3​[ex−11​]dx=α−loge​2, then α3\alpha^3α3 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Simplify the integrand

We have

[1ex−1]=[e1−x].\left[\frac{1}{e^{x-1}}\right] = \left[e^{1-x}\right].[ex−11​]=[e1−x].

So the integral is

I=∫0e3[e1−x] dx.I=\int_0^{e^3} \bigl[ e^{1-x} \bigr] \, dx.I=∫0e3​[e1−x]dx.
  1. Find where the greatest integer value changes

Since e1−xe^{1-x}e1−x is a decreasing function of xxx:

  • At x=0x=0x=0, e1−x=ee^{1-x}=ee1−x=e, so [e1−x][e^{1-x}][e1−x] can take values 2,1,02,1,02,1,0 on the interval.
  • [e1−x]≥2[e^{1-x}]\ge 2[e1−x]≥2 when e1−x≥2  ⟺  1−x≥ln⁡2  ⟺  x≤1−ln⁡2.e^{1-x}\ge 2 \iff 1-x\ge \ln 2 \iff x\le 1-\ln 2.e1−x≥2⟺1−x≥ln2⟺x≤1−ln2.
  • Also, [e1−x]≥1[e^{1-x}]\ge 1[e1−x]≥1 when e1−x≥1  ⟺  1−x≥0  ⟺  x≤1.e^{1-x}\ge 1 \iff 1-x\ge 0 \iff x\le 1.e1−x≥1⟺1−x≥0⟺x≤1.
  • For x>1x>1x>1, we have 0<e1−x<10<e^{1-x}<10<e1−x<1, hence [e1−x]=0.[e^{1-x}] = 0.[e1−x]=0.

Thus,

[e1−x]={2,0≤x≤1−ln⁡2,1,1−ln⁡2<x≤1,0,1<x≤e3.[e^{1-x}] = \begin{cases} 2, & 0\le x\le 1-\ln 2,\\[4pt] 1, & 1-\ln 2 < x\le 1,\\[4pt] 0, & 1 < x\le e^3. \end{cases}[e1−x]=⎩⎨⎧​2,1,0,​0≤x≤1−ln2,1−ln2<x≤1,1<x≤e3.​
  1. Evaluate the integral piecewise

So,

I=∫01−ln⁡22 dx+∫1−ln⁡211 dx+∫1e30 dx.I=\int_0^{1-\ln 2} 2\,dx + \int_{1-\ln 2}^{1} 1\,dx + \int_1^{e^3} 0\,dx.I=∫01−ln2​2dx+∫1−ln21​1dx+∫1e3​0dx.

Therefore,

I=2(1−ln⁡2)+(1−(1−ln⁡2)).I = 2(1-\ln 2) + (1-(1-\ln 2)).I=2(1−ln2)+(1−(1−ln2)).

Now simplify:

I=2−2ln⁡2+ln⁡2=2−ln⁡2.I = 2-2\ln 2 + \ln 2 = 2-\ln 2.I=2−2ln2+ln2=2−ln2.
  1. Compare with the given form

Given

∫0e3[1ex−1]dx=α−log⁡e2.\int_0^{e^3}\left[\frac{1}{e^{x-1}}\right]dx = \alpha - \log_e 2.∫0e3​[ex−11​]dx=α−loge​2.

Since log⁡e2=ln⁡2\log_e 2 = \ln 2loge​2=ln2, we get

α−ln⁡2=2−ln⁡2.\alpha - \ln 2 = 2 - \ln 2.α−ln2=2−ln2.

Hence,

α=2.\alpha = 2.α=2.
  1. Compute α3\alpha^3α3
α3=23=8.\alpha^3 = 2^3 = 8.α3=23=8.

Final Answer:

8\boxed{8}8​
  1. Comparison with stored correct answer

Stored correct answer = 888.

Our derived answer also equals 888, so they agree.

Next

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