A) \[\pi \]
B) \[\pi /2\]
C) \[1\]
D) 0
Correct Answer: D
Solution :
Let \[I=\int_{-\pi /2}^{\pi /2}{\{f(x)+f(-x)\}\{g(x)-g(-x)\}dx}\] Again,let \[h(x)=\{f(x)+f(-x)\}\{g(x)-g(-x)\}\] \[\Rightarrow \] \[h(-x)=\{f(-x)+f(x)\}\{g(-x)-g(x)\}\] \[\Rightarrow \] \[h(-x)=-h(x)\] Hence,\[h(x)\]is an odd function. \[\therefore \] \[I=0\]You need to login to perform this action.
You will be redirected in
3 sec