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JEE Mains · Maths · STD 12 - 7.2 definite integral

Let \(f: R \rightarrow R\) be a continuous function such that \(f(x)+f(x+1)=2,\) for all \(x \in R\). If \(I _{1}=\int_{0}^{8} f( x ) d x\) and \(I _{2}=\int_{-1}^{3} f( x ) d x ,\) then the value of \(I _{1}+2 I _{2}\) is equal to...........

  1. A \(25\)
  2. B \(16\)
  3. C \(32\)
  4. D \(40\)
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Answer & Solution

Correct Answer

(B) \(16\)

Step-by-step Solution

Detailed explanation

\(f(x)+f(x+1)=2\) \(\Rightarrow f( x )\) is periodic with period \(=2\) \(I_{1}=\int_{0}^{8} f(x) d x=4 \int_{0}^{2} f(x) d x\) \(=4 \int_{0}^{1}(f(x)+f(1+x)) d x=8\) Similarly \(I _{2}=2 \times 2=4\) \(I _{1}+2 I _{2}=16\)
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