11.br=1/r+1
I=\int_{0}^{\pi }{(sinx/2+sin2x/3+sin3x/4.....)}=\prod_{0}^{\pi }{(-cosx/2-cos)2x/3-cos3x/4....)}
=2(1/1.2+1/3.4+1/5.6 ....)=2ln2
2nd one can be solved by takin arbitrary values for a0,a1...and also for 'n'.
f(x)=a_0+a_1cosx+a_2cos2x+a_3cos3x+...+a_ncosnx
g(x)=b_1sinx+b_2sn2x+b_3sin3x+..+b_nsinnx
Q1 If br=1r+1,then \lim_{n\rightarrow \infty} \int_{0}^{\pi }{g(x)dx}
Q2 \sum_{k=0}^{n}{\int_{0}^{2\pi }{f(x).(cos kx)dx}}
Q3 \sum_{k=0}^{n}{\int_{0}^{2\pi }{g(x).(cos kx)dx}}
-
UP 0 DOWN 0 1 4
4 Answers
113.taking into account
\int_{0}^{2\pi }{sinnxcosmx}=0
and\int_{0}^{2\pi }{sinx}=0
we get ans as 0.
1i think rahul what u did was wrong in post 1......
if Br = 1/r+1 then
I = \int_{0 }^{\Pi }({\sin x/2}+\sin 2x/3 + sin 3x/4....+sin nx/n+1)
not what u wrote
