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Integral Calculus

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11 Feb 2008 09:21:20 IST

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Evaluate ₀ |sin1999x| - |sin2000x| dx

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abhishek sinha

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11 Feb 2008 10:10:16 IST

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edited !!!

Hari Shankar

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11 Feb 2008 10:16:48 IST

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The answer is incorrect bcos my qn is typed in wrongly.

But, cud qns by engg stud/alumni be left to the aspirants to solve. I am not trying to find out how to do the problem

abhishek sinha

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11 Feb 2008 10:19:03 IST

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edited !!!!!!!!

Hari Shankar

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11 Feb 2008 10:20:42 IST

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i had typed the qn wrong. But, i again request you to leave it alone sir.

Rahul Raghavendra

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11 Feb 2008 10:20:56 IST

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is it

2/1999*2000

Hari Shankar

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11 Feb 2008 10:26:17 IST

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i am sorry raul, i've revised the qn

Rahul Raghavendra

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11 Feb 2008 10:34:38 IST

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is it 1/1000

Hari Shankar

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11 Feb 2008 10:37:15 IST

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It is better you give your working raul.

Sairam

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12 Feb 2008 12:32:50 IST

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_{downld img for better view}

Hari Shankar

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12 Feb 2008 13:20:56 IST

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elasti, can i trouble you to write a more complete soulution. You are on the rt track. It will help others and me too.

Sairam

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12 Feb 2008 17:56:30 IST

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as usual[ ] stands for mod

we know,

[sinx] has period pi, which can easily be inferred from the graph of[sinx]

also if f(x)has period p,

f(ax) has period p/[a]

use these both to split the integrals

_{[0 ]}

^{[pi ]} sin[1999x]=1999_{[0 ]}

^{[pi/1999 ]} sin[1999x]

now sin1999x is positive in the interval (o,pi/1999)

we get 1999*_{[ 0]}

^{[pi/1999 ]} sin1999x

Integr. 1999*{-cos1999x/1999} between the limits 0- pi/1999

it turns out to be 2.

Now the other integral can be similiarly dealt wid by splittin at integral multiples of pi/2000.

The other integral is also 2.

so diff.=0.

Sairam

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12 Feb 2008 18:12:01 IST

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after doin this problem

I figured it out dat

_{[ 0]}

^{[pi ]} [sinnx] is always 2!!!

Well one way of graphically interpreting this is as n attains higher values the

graph becomes narrower .

P.S :forgive me if the graph is bad, iam bad at drawing with paint

Hari Shankar

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12 Feb 2008 19:28:34 IST

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The key to the solution as elastiboysai pointed out is that |sinx| has a period of

So ₀ |sin(nx)| dx = 1/n ₀ⁿ |sint| dt where t = nx

Now since |sint| has a period of

, the integral can be written as

[0,

]

and ₀ sint dt = 2.

Hence the required integral works out to 0.

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