Let the function f:[0,2] → ℝ be defined as
f(x)= e^(min {x², x-[x]},) x ∈[0,1) ; e^([x-log ₑ x]), x ∈[1,2]
where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫0² x f(x) d x is :

Solution

Maths sir · 3 steps

Settle each greatest-integer bracket on its own interval first, and both pieces of f become easy.

1. Piece on [0, 1). Here [x] = 0, so x - [x] = x. For 0 ≤ x < 1 we have x² ≤ x, so the minimum is x² and f(x) = e^(x²). With t = x², ∫₀¹ x e^(x²) dx = 1/2[e^(x²)]₀¹ = (e-1)/2.
2. Piece on [1, 2]. Let g(x) = x - ln x. Then g'(x) = 1 - 1/x ≥ 0, so g is increasing, from g(1) = 1 to g(2) = 2 - ln 2 ≈ 1.31. So [g(x)] = 1 on the whole interval, f(x) = e, and ∫₁² e x dx = e × (4-1)/2 = 3e/2.

   ∫₀² x f(x) dx = (e-1)/2 + 3e/2 = 2e - 1/2

3. So. The integral equals 2e - 1/2, which is option (b).

The idea to keep. With [ ] or min inside an integrand, split at the break points and find which expression wins on each piece. For a bracket like [x - ln x], find the range of the inside using its derivative.

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