Let f: ℝ → ℝ be a function defined as
f(x)=a sin (π[x]/2)+[2-x], a ∈ ℝ where [t] is the greatest integer less than or equal to t. If lim _(x → -1) f(x) exists, then the value of ∫₀⁴ f(x) d x is equal to

Solution

Maths sir · 4 steps

Match the two one-sided limits at -1 to get a, then integrate f one unit strip at a time.

1. Left of minus one. For x → -1⁻, [x] = -2, so sin(-π) = 0. Also 2 - x → 3⁺, so [2-x] = 3. The left-hand limit is 3.
2. Right of minus one. For x → -1⁺, [x] = -1, so sin(-π/2) = -1. Also 2 - x → 3⁻, so [2-x] = 2. The right-hand limit is 2 - a.
3. Find a. The limit exists, so 2 - a = 3, which gives a = -1. Now f(x) = -sin((π [x])/2) + [2-x], and f is constant on each open unit interval. The integer points do not change the integral.

   Interval	-sin(π[x]/2)	[2-x]	f(x)
   (0, 1)	-sin 0 = 0	1	1
   (1, 2)	-sin π/2 = -1	0	-1
   (2, 3)	-sin π = 0	-1	-1
   (3, 4)	-sin 3π/2 = 1	-2	-1

4. So. Each strip has width 1, so the integral is 1 - 1 - 1 - 1 = -2. That is option (b).

The idea to keep. To integrate a greatest-integer function, split at the integers. On each piece the function is a constant, so the integral is just a sum of rectangles.

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