Consider the integral
I = ∫₀¹⁰ [x]e^([x])/eˣ⁻¹dx,
where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :

Solution

Maths sir · 4 steps

Break the integral at every integer: on each piece [x] is a constant, and the rest is a plain exponential.

1. Split at integers. On [k, k+1) we have [x] = k, so the integrand becomes k eᵏe⁻⁽ˣ⁻¹⁾ = k eᵏ⁺¹⁻ˣ. Split I into 10 pieces, for k = 0, 1, …, 9.

   ∫ₖᵏ⁺¹ k eᵏ⁺¹⁻ˣ dx = k[-eᵏ⁺¹⁻ˣ]ₖᵏ⁺¹ = k(e - 1)

2. Same value each time. Every piece gives e - 1 times its integer k. The exponential part does not depend on k, because k+1-x always runs from 1 down to 0.
3. Add the pieces. I = (e-1)(0 + 1 + 2 + … + 9) = (e-1)×(9 × 10)/2 = 45(e-1).
4. So. I = 45(e-1), which is option (a).

The idea to keep. With [x] inside an integral, break the limits at every integer. On each piece [x] is a constant, and the pieces often repeat, so you end up summing a simple series.

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