Let [t] denote the greatest integer less than or equal to t. Then the value of the integral ∫₋₃¹⁰¹([sin (π x)]+e^([cos (2 π x)])) d x is equal to

Solution

Maths sir · 3 steps

Both parts are periodic, so find each one over a single period and multiply by the number of periods.

1. Spot the periods. [sin(π x)] has period 2 and e^([cos(2π x)]) has period 1. The interval from -3 to 101 has length 104, which is 52 periods of the first part and 104 periods of the second.

   Part and interval	Value of the bracket	Integrand	Integral over one period
   [sin π x] on (0, 1)	0 < sin π x ≤ 1, so 0 (except at the single point x = 1/2)	0	0 + (-1)(1) = -1
   [sin π x] on (1, 2)	-1 ≤ sin π x < 0, so -1	-1	(included in the row above)
   e^([cos 2π x]) on (0, 1/4) ∪ (3/4, 1)	0 < cos 2π x < 1, so 0	e⁰ = 1	1/2(1) + 1/2(1/e) = 1/2 + 1/2e
   e^([cos 2π x]) on (1/4, 3/4)	-1 ≤ cos 2π x < 0, so -1	e⁻¹	(included in the row above)

2. Multiply by periods. The first part gives 52 × (-1) = -52. The second part gives 104 × (1/2 + 1/2e) = 52 + 52/e.

   -52 + 52 + 52/e = 52/e

3. So. The integral equals 52/e, which is option (b).

The idea to keep. When an integrand is periodic, integrate over one period and multiply. Inside one period, find where the bracket is constant. Points where it jumps do not change the area.

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