If f "and" g are continuous functions in [0, 1] satisfying f(x)=f(a-x) and g(x)+g (a-x)=a, then  ∫ 0 a f(x). g(x) dx is equal to

(A) a/2 

(B) a/2 ∫ 0^a f(x)  dx

(C) ∫ 0^a f(x) dx  

(D) a∫ 0^a f(x) dx

This question is similar to Ex 7.11, 19 - Chapter 7 Class 12 - Integrals




Question 5 If 𝑓 "and" 𝑔 are continuous functions in [0, 1] satisfying 𝑓(𝑥)=𝑓(𝑎−𝑥) and 𝑔(𝑥)+𝑔 (𝑎−𝑥)=𝑎, then ∫1_0^𝑎▒〖𝑓(𝑥). 𝑔(𝑥)〗 𝑑𝑥 is equal to 𝑎/2 (B) 𝑎/2 ∫1_0^𝑎▒〖𝑓(𝑥) 𝑑𝑥〗 (C) ∫1_0^𝑎▒〖𝑓(𝑥) 𝑑𝑥〗 (D) 𝑎∫1_0^𝑎▒〖𝑓(𝑥) 𝑑𝑥〗 Let 𝑰 =∫_𝟎^𝒂▒𝒇(𝒙) 𝒈(𝒙) 𝒅𝒙 Using g(𝑥)+𝑔(𝑎−𝑥)=𝑎 I =∫_0^𝑎▒𝑓(𝑥) [𝑎−𝑔(𝑎−𝑥)] 𝑑𝑥 I = ∫_0^𝑎▒[𝑎.𝑓(𝑥)−𝑓(𝑥)𝑔(𝑎−𝑥)] 𝑑𝑥 𝑰 =𝒂∫_𝟎^𝒂▒〖𝒇(𝒙)𝒅𝒙−∫_𝟎^𝒂▒〖𝒇(𝒙) 𝒈(𝒂−𝒙) 〗〗 𝒅𝒙 I =𝑎∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥−∫_𝟎^𝒂▒〖𝒇(𝒂−𝒙) 𝒈(𝒂−(𝒂−𝒙)) 〗〗 𝑑𝑥 I =𝑎∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥−∫_0^𝑎▒〖𝒇(𝒂−𝒙) 𝑔(𝑥) 〗〗 𝑑𝑥 Using 𝑓(𝑥)=𝑓(𝑎−𝑥) I =𝑎∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥−∫_0^𝑎▒〖𝒇(𝒙) 𝑔(𝑥) 〗〗 𝑑𝑥 I =𝑎∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥−𝐈〗 I +I=a∫_0^𝑎▒𝑓(𝑥)𝑑𝑥 2I=a∫_0^𝑎▒𝑓(𝑥)𝑑𝑥 𝐈=𝐚/𝟐 ∫_𝟎^𝒂▒𝒇(𝒙)𝒅𝒙 ∴ ∫_0^𝑎▒〖𝑓(𝑥) 𝑔(𝑥) 〗 𝑑𝑥=2∫_0^𝑎▒𝑓(𝑥)𝑑𝑥 Hence Proved So, the correct answer is (b)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.