lim n--->oo n^2∫(2007 sinx+2008 cosx)|x| (integrate from -1/n to 1/n)
Pranjal Joshi , 5 Years ago
Grade 12th pass
IIT JEE Entrance Exam> lim n--->oo n^2∫(2007 sinx+2008 cosx)|x| ...
1 Answers
Aditya Gupta
Last Activity: 5 Years ago
hello pranjal, this is an ez ques.
first put y= 1/n
so limit becomes lim y--->0 ∫(2007 sinx+2008 cosx)|x|dx/y^2 (integrate from – y to y)
now, note that ∫(2007 sinx+2008 cosx)|x|dx= ∫2007 sinx*|x|dx from – y to y + ∫2008 cosx*|x|dx from – y to y.
now ∫2007 sinx*|x|dx from – y to y= 0 since 2007sinx*|x| is an odd function.
also, since 2008 cosx*|x| is an even function, we can write ∫2008 cosx*|x|dx from – y to y= 2*∫2008 cosx*|x|dx from 0 to y= 2*∫2008 cosx*xdx from 0 to y (as x is positive now in the changed limits)
so, our limit becomes
lim y--->0 2*[∫2008 cosx*xdx from 0 to y.]/y^2
= 2*2008*lim y--->0 [∫cosx*xdx from 0 to y.]/y^2
now this above limit is 0/0 form, so apply hospital rule by diff wrt y and using the formula of differentiation under the integral sign.
we have 2*2008*lim y--->0 ycosy/2y
= 2008*lim y--->0 cosy
= 2008
kindly approve :=)
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