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Differential Calculus

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30 Jan 2007 13:59:15 IST

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1143

SOLVE FOR A SALUTE!!!

None

Find the value of

_{[ n]}

_{[ infinity ]} _{[ r=1]}

^{[ n]} n²/(r³+n³)

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Comments (18)

vineet

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30 Jan 2007 14:18:05 IST

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is it 0

CyBorG

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30 Jan 2007 14:20:19 IST

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Well the ans is not 0.Think in a different way.

$h0cK3r

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30 Jan 2007 14:29:17 IST

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oh it is limit of sum=integral 0 to 1 1/(1+x^3)

$h0cK3r

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30 Jan 2007 15:07:25 IST

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the value coming out is 0.83565

akhil

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30 Jan 2007 15:15:23 IST

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=(1/(r/n)^3 +1) / n

let r/n=x
f(x)=1/(x^3+1)

So the given ques becomes:

integration from 0 to 1 f(x).

CyBorG

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30 Jan 2007 15:23:42 IST

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I know the ans .Whoever answers it first(correctly) is going to get a SALUTE!!!

akhil

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30 Jan 2007 15:28:39 IST

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integrate it using partial fractions.

writting 5^1/2 as c.

The ans is (ln2)/3 +c*ln((7-3c)/2)/6

CyBorG

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30 Jan 2007 18:46:53 IST

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Well it is not the correct ans!

vishak p

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24 Feb 2007 14:07:58 IST

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adarsh, this uses definte integration ! take 1/n common, then there shall be a function in r/n.take that to be x and decide the limits !

vishak p

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24 Feb 2007 14:15:26 IST

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_{[ n][ infinity ]} _{[ r=1]}

^{[ n]} n²/(r³+n³)

ok, now this is equal to ::

_{[ n][ infinity ]} _{[ r=1]}

^{[ n]}1/n [ n³/ r³ + n³]

= _{n][ infinity ]} _{[ r=1]}

^{[ n]}1/n [ 1/ { (r/n)³+ 1 ]

take r/n = x => limts become 0 to 1

the above expression is equal to

_{[0 ]}

^{[1 ]} (dx / x³+1)

solve and get the answer !

amaron

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24 Feb 2007 14:42:32 IST

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Is it 1/2 +1/3 + 1/4 +........

vishak p

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24 Feb 2007 14:43:27 IST

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what is the answer ??

CyBorG

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24 Feb 2007 14:46:22 IST

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Check my answer in my post in integral calculus.

vishak p

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24 Feb 2007 14:48:11 IST

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is my method correct ??

CyBorG

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24 Feb 2007 14:49:45 IST

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Your method is perfect.But there are chances of making mistakes in furthur steps after integrating.

vishak p

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24 Feb 2007 14:49:50 IST

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yes, i am also getting the same answer !(log2)/3+pi/(3sqrt3)

vishak p

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24 Feb 2007 14:52:30 IST

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the answer's roughly 0.84

Divya Alok Gupta

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24 Feb 2007 20:55:54 IST

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Yes it is

dx/x^3 +1 =

dx/(x+1)(x^2-x+1)

On using partial fractions

we get

1/3

dx/(x+1) - 1/3

x-2dx/(x^2 - x +1)

Now u can solve this easily

right!!!!

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