Integral Calculus Question 145
Question: If $ I_1=\int\limits_0^{\pi }{xf({{\sin }^{3}}x+{{\cos }^{2}}x)dx} $ and $ I_2=\pi \int\limits_0^{\pi /2}{f({{\sin }^{3}}x+{{\cos }^{2}}x)dx} $ , then
Options:
A) $ I_1=2I_2 $
B) $ 2I_1=I_2 $
C) $ I_1=I_2 $
D) $ I_1+I_2=0 $
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Answer:
Correct Answer: C
Solution:
[c] Consider $ I_1=\int\limits_0^{\pi }{xf[ {{\sin }^{3}}x+{{\cos }^{2}}x ]}dx $ $ =\int\limits_0^{\pi }{(\pi -x)f[ {{\sin }^{3}}(\pi -x)+{{\cos }^{2}}(\pi -x) ]dx} $ $ =\int\limits_0^{\pi }{(\pi -x)f[ {{\sin }^{3}}x+{{\cos }^{2}}x ]dx} $ $ =\int\limits_0^{\pi }{\pi f({{\sin }^{3}}x+{{\cos }^{2}}x)dx-\int\limits_0^{\pi }{xf({{\sin }^{3}}x+{{\cos }^{2}}x)dx}} $ $ =\int\limits_0^{\pi }{\pi f({{\sin }^{3}}x+{{\cos }^{2}}x)dx-I_1} $
$ \Rightarrow 2I_1=\int\limits_0^{\pi }{\pi f({{\sin }^{3}}x+{{\cos }^{2}}x)dx} $ $ 2I_1=2\pi \int\limits_0^{\pi /2}{f({{\sin }^{3}}x+{{\cos }^{2}}x)dx} $
$ \Rightarrow I_1=\pi \int\limits_0^{\pi /2}{f({{\sin }^{3}}x+{{\cos }^{2}}x)dx} $ $ I_1=I_2 $ (By definition of $ I_2 $ )
BETA
