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JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Definite Integrals

Let [ • ] denote the greatest integer function. If

\int_{0}^{e^{3}} [\frac{1}{e ^{x - 1}}] dx = α - lo g_{e} 2

, then

α^{3}

is equal to ______.

Answer: 8

Solution

Let

f (x) = \frac{1}{e ^{x - 1}} = e^{1 - x}

.

At

x = 0

:

f (0) = e \approx 2.71

, so

[f (0)] = 2

.

Find where

f (x) = 2

:

e^{1 - x} = 2 \Rightarrow 1 - x = ln 2 \Rightarrow x = 1 - ln 2

.

Find where

f (x) = 1

:

e^{1 - x} = 1 \Rightarrow x = 1

.

At

x = e^{3}

:

f (e^{3}) = e^{1 - e^{3}} \approx 0

, so

[f (e^{3})] = 0

.

The integral splits:

I = \int_{0}^{1 - l n 2} 2 d x + \int_{1 - l n 2}^{1} 1 d x + \int_{1}^{e^{3}} 0 d x

= 2 (1 - ln 2) + (1 - (1 - ln 2)) + 0

= 2 - 2 ln 2 + ln 2 = 2 - ln 2

Comparing with

α - ln 2

, we get

α = 2

.

Therefore,

α^{3} = 8

.

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Integer answer

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