Let [.] denote the greatest integer function. If ∫0^(e³)[1/eˣ⁻¹] d x=α-log ₑ 2, then α³ is equal to _________.

Solution

Maths sir · 3 steps

Rewrite the integrand as e¹⁻ˣ and find where it crosses 2 and 1. Those are your break points.

1. Simplify first. 1/eˣ⁻¹ = e¹⁻ˣ. This is a decreasing function. It starts at e ≈ 2.718 when x = 0 and keeps falling.
2. Find break points. e¹⁻ˣ = 2 when x = 1 - ln 2, and e¹⁻ˣ = 1 when x = 1. For x > 1 the value is below 1, so the greatest integer is 0 all the way up to e³.

   Interval	Value of e¹⁻ˣ	[e¹⁻ˣ]	Contribution
   0 ≤ x < 1 - ln 2	between 2 and e	2	2(1 - ln 2)
   1 - ln 2 < x ≤ 1	between 1 and 2	1	1 × ln 2
   1 < x ≤ e³	less than 1	0	0

   ∫₀^(e³)[e¹⁻ˣ]dx = 2(1-ln 2) + ln 2 = 2 - ln 2

3. So. Comparing with α - logₑ 2 gives α = 2, so α³ = 8.

The common slip: 2. Stopping at α and reporting 2 misses the last step. The question asks for α³, which is 8.

The idea to keep. With a greatest integer function, find the points where the inside crosses each integer. Between those points the integrand is a constant, so each piece is a rectangle.

Answer from the NTA final answer key. Solution by PrepWiser.