Question:
Let \(f : R \to R\) be a continuous function such that \(f(x) + f(x+1) = 2\), for all \(x \in R\). If \(I_1 = \int\limits_0^8 f(x)dx\) and \(I_2 = \int\limits_{-1}^3 f(x)dx\), then the value of \(I_1 + 2I_2\) is equal to ____________.
Let \(f : R \to R\) be a continuous function such that \(f(x) + f(x+1) = 2\), for all \(x \in R\). If \(I_1 = \int\limits_0^8 f(x)dx\) and \(I_2 = \int\limits_{-1}^3 f(x)dx\), then the value of \(I_1 + 2I_2\) is equal to ____________.
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From the functional relation f has period 2 and \(\int_a^{a+2} f = 2\) for every a.
Updated On: Nov 4, 2026
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Correct Answer: 16
Solution and Explanation
\(f(x) + f(x+1) = 2 \Rightarrow f(x+1) + f(x+2) = 2\), so \(f(x+2) = f(x)\): f has period 2.
For any a: \(\int_a^{a+2} f(x)dx = \int_a^{a+1} f(x)dx + \int_a^{a+1} f(x+1)dx = \int_a^{a+1}\left(f(x)+f(x+1)\right)dx = 2\).
\(I_1 = \int_0^8 f = 4 \times 2 = 8\). \(I_2 = \int_{-1}^{3} f = 2 \times 2 = 4\).
\(I_1 + 2I_2 = 8 + 8 = 16\).
Answer: 16
For any a: \(\int_a^{a+2} f(x)dx = \int_a^{a+1} f(x)dx + \int_a^{a+1} f(x+1)dx = \int_a^{a+1}\left(f(x)+f(x+1)\right)dx = 2\).
\(I_1 = \int_0^8 f = 4 \times 2 = 8\). \(I_2 = \int_{-1}^{3} f = 2 \times 2 = 4\).
\(I_1 + 2I_2 = 8 + 8 = 16\).
Answer: 16
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