Definite Integration Question 12
Question: $ [ \sum\limits _{n=1}^{10}{\int _{-2n-1}^{2n}{{{\sin }^{27}}xdx}} ]+[ \sum\limits _{n=1}^{10}{\int_2n^{2n+1}{{{\sin }^{27}}xdx}} ] $ equals
[MP PET 2002]
Options:
A) $ 27^{2} $
B) $ -54 $
C) 36
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
$ \sum\limits _{n=1}^{10}{\int _{-2n-1}^{2n}{{{\sin }^{27}}xdx+}}\sum\limits _{n=1}^{10}{\int_2n^{2n+1}{{{\sin }^{27}}xdx}} $
In the first summation put $ x=-t $
$ \sum\limits _{n=1}^{10}{{{\sin }^{27}}(-t)(-dt)+\sum\limits _{n=1}^{10}{\int_2n^{2n+1}{{{\sin }^{27}}xdx}}} $
$ =-\sum\limits _{n=1}^{10}{\int_2n^{2n+1}{{{\sin }^{27}}}xdx+}\sum\limits _{n=1}^{10}{\int_2n^{2n+1}{{{\sin }^{27}}}xdx=0} $ .
BETA
