Integral Calculus Question 161
Question: $ [ \sum\limits_{n=1}^{10}{\int\limits_{-2n-1}^{-2n}{{{\sin }^{27}}xdx}} ]+[ \sum\limits_{n=1}^{10}{\int\limits_2n^{2n+1}{{{\sin }^{27}}}xdx} ]= $
Options:
A) $ 27^{2} $
B) $ -54 $
C) $ 54 $
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
[d] General term of the series $ \sum\limits_{n=1}^{10}{\int\limits_{-2n-1}^{-2n}{{{\sin }^{27}}xdx}} $ is $ I_1=\int\limits_{-2n-1}^{-2n}{{{\sin }^{27}}xdx=\int\limits_{2n+1}^{2n}{{{\sin }^{27}}(-x)(-dx)}} $ $ =-\int\limits_2n^{2n+1}{{{\sin }^{27}}xdx=-I_2} $ Where $ I_2 $ is general term of series $ \sum\limits_{n=1}^{10}{\int\limits_2n^{2n+1}{{{\sin }^{27}}xdx}} $ So $ I_1+I_2=0 $ for all n
BETA
