Definite Integration Question 102
Question: If $ g(x)=\int_0^{x}{{{\cos }^{4}}tdt,} $ then $ g(x+\pi ) $ equals
[IIT 1997 Re-Exam; DCE 2001; UPSEAT 2001; Pb. CET 2002]
Options:
A) $ g(x)+g(\pi ) $
B) $ g(x)-g(\pi ) $
C) $ g(x)g(\pi ) $
D) $ g(x)/g(\pi ) $
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Answer:
Correct Answer: A
Solution:
$ g(x+\pi )=\int_0^{x+\pi }{{{\cos }^{4}}tdt=\int_0^{\pi }{{{\cos }^{4}}tdt+\int _{\pi }^{x+\pi }{{{\cos }^{4}}tdt}}} $
$ =g(\pi )+f(x) $
$ f(x)=\int_0^{x}{{{\cos }^{4}}udu=g(x)} $ , $ (\because t=\pi +u) $
$ \therefore g(x+\pi )=g(x)+g(\pi ) $ .
BETA
