JEE Mains · Maths · STD 12 - 7.2 definite integral
If \(f(a+b+1-x)=f(x),\) for all \(x,\) where \(a\) and \(b\) are fixed positive real numbers, then \(\frac{1}{a+b} \int\limits_{a}^{b} x(f(x)+f(x+1)) d x\) is equal to
- A \(\int\limits_{a+1}^{b+1} f(x) d x\)
- B \(\int\limits_{a+1}^{b+1} f(x+1) d x\)
- C \(\int\limits_{a+1}^{b-1} f(x+1) d x\)
- D \(\int\limits_{a-1}^{b-1} f(x) d x\)
Answer & Solution
Correct Answer
(A) \(\int\limits_{a+1}^{b+1} f(x) d x\)
Step-by-step Solution
Detailed explanation
\(f(x+1)=f(a+b-x)\) \(I=\frac{1}{(a+b)} \int_{a}^{b} x(f(x)+f(x+1) d x \ldots(1)\) \(I=\frac{1}{(a+b)} \int_{a}^{b}(a+b-x)(f(x+1)+f(x)) d x \ldots(2)\) from \(( 1)\) and \(( 2)\) \(2 \mathrm{I}=\int_{a}^{b}(\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}+1) \mathrm{d} \mathrm{x}\)…
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