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`        lim n--->oo n^2∫(2007 sinx+2008 cosx)|x| (integrate from -1/n to 1/n)`
one month ago

## Answers : (1)

1673 Points
```							hello pranjal, this is an ez ques.first put y= 1/nso 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 becomeslim y--->0 2*[∫2008 cosx*xdx from 0 to y.]/y^2= 2*2008*lim y--->0 [∫cosx*xdx from 0 to y.]/y^2now 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= 2008kindly approve :=)
```
one month ago
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• Test paper with Video Solution
• Mind Map
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• Discussion Forum
• Previous Year Exam Questions