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Integral Calculus

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29 Mar 2009 19:04:44 IST

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292

definite integral

None

definite integral

Thanks , First one is clear .

But still not clear in the 2nd one .

What i did is "

0 to pi/2 , if i put the minimum value then cos x is cos 0 = 1

and from pi/2 to pi , if i put the minium value then cos pi/2 =0 ..

tell me where im doing themistake . Thanks

I am able to get the result , i m getting 9/4 but the answer is 11/4 .

And one more

Thanks

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Comments (6)

Sahil Gupta

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29 Mar 2009 19:18:14 IST

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the expression is equal to zero at 0,1,2 and if you use differential calculus the ex. is +ve from 0 to 1 and 2 to 3 and -ve from 1 to 2. So the area comes out to be 1/4 -(-1/4) + 9/4 = 11/4

Sahil Gupta

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29 Mar 2009 19:20:35 IST

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The answer is for the second one.

Madmax

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29 Mar 2009 19:22:12 IST

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The eqn is +ve in 0 to 1, -ve in 1 to 2 and +ve again in 2 to 3

Keep this region and U get integral for 0 to 1 minus integral from 1 to 2 plus integral from 2 to 3 of x^3-3x^2+2x

The answer is definitley 11/4 (The book is right)

cheers!!

A$PIRing HiGh - dEv aDitYa

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29 Mar 2009 19:22:43 IST

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Second one. As cos x is +ve in first quad. and max value is +1, So from 0 to pi/2 integration results 0. and from pi/2 to pi, cos x is -ve and min is -1, it results integratin of -1 from pi/2 to pi, So, ans is -pi/2. Hope it's correct

Madhulika

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29 Mar 2009 21:24:15 IST

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Thanks , First one is clear .

But still not clear in the 2nd one .

What i did is "

0 to pi/2 , if i put the minimum value then cos x is cos 0 = 1

and from pi/2 to pi , if i put the minium value then cos pi/2 =0 ..

So I_{1= pi/2}

_{and I2=0 ( As cos pi/2 0 )}

_{Please correct me .}

tell me where im doing themistake . Thanks

Tejal Dhabhai

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Joined: 27 Mar 2009

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29 Mar 2009 21:37:32 IST

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first u find d factors of f(x). d factors vil b 0, 1, 2. then d functn wil b ientegratd frm 0-1, 1-2, 2-3 one by one in d fst question. so d ans is (1/4)-(-1/4)+ 9/4= 11/4.2nd quest isnt clearPlz rate d ans.

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