Let f : R → R be a continuous function such that f(x) + f(x + 1) = 2, for all x∈R.
If I₁ = ∫₀⁸ f(x)dx and I₂ = ∫₋₁³ f(x)dx, then the value of I₁ + 2I₂ is equal to ____________.

Solution

Maths sir · 4 steps

Shift the given relation by 1 and you will see that f repeats every 2 units. Then just count periods.

1. Find the period. Replace x by x+1 to get f(x+1) + f(x+2) = 2. Subtract this from f(x) + f(x+1) = 2 to get f(x+2) = f(x). So f is periodic with period 2.
2. One period. Integrate the given relation from 0 to 1. The second term becomes ∫₁² f(x) dx, so the two pieces join into one full period of length 2.

   ∫₀¹ f(x) dx + ∫₁² f(x) dx = ∫₀² f(x) dx = ∫₀¹ 2 dx = 2

3. Count periods. [0, 8] has length 8, which is 4 periods, so I₁ = 4 × 2 = 8. [-1, 3] has length 4, which is 2 periods, so I₂ = 2 × 2 = 4. For a periodic function, the integral over any full period is the same wherever the period starts.
4. So. I₁ + 2I₂ = 8 + 2 × 4 = 16.

The common slip: 32. Taking the integral over a length of 1 as 2, instead of the integral over a full period of length 2, gives I₁ = 16 and I₂ = 8, so the total comes out as 32.

The idea to keep. When f(x) + f(x+a) is constant, f has period 2a. Integrate the relation over a length a to get the integral over one period, then count how many periods each interval holds.

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