WBJEE · Maths · Definite Integration
Let \(\mathrm{f}\) be derivable in \([0,1]\), then
- A there exists \(c \in(0,1)\) such that \(\int_0^c f(x) d x=(1-c) f(c)\)
- B there does not exist any point \(d \in(0,1)\) for which \(\int_0^d f(x) d x=(1-d) f(d)\)
- C \(\int_0^c f(x) d x\) does not exist, for any \(c \in(0,1)\)
- D \(\int_0^c f(x) d x\) is independent of \(c, c \in(0,1)\)
Answer & Solution
Correct Answer
(A) there exists \(c \in(0,1)\) such that \(\int_0^c f(x) d x=(1-c) f(c)\)
Step-by-step Solution
Detailed explanation
Let \(g(x)=x \int_0^x f(t) d t-\int_0^x f(t) d t\) Now, \(g(0)=0\) \(g(1)=0\) By Rolle's Theorem \(g^{\prime}(x)=0\) for some \(x \in(0,1)\) \(g^{\prime}(x)=x f(x)+\int_0^x f(t) d t-f(x)=0\)
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