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Integral Calculus

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14 Mar 2010 13:23:53 IST

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if line y=mx+2 cuts parabola 2y=x^2 at points (x1,y1) & (x2,y2) where x1<x2 thn value of >

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if line y=mx+2 cuts parabola 2y=x^2 at points (x1,y1) & (x2,y2) where x1<x2 thn value of m for which integral mx+2+-(x^2)/2 dx upr limit x2 lower limit x1value of m is minimum-?

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Deepak Aggarwal

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Joined: 9 May 2009

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15 Mar 2010 16:46:21 IST

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Find the intersection of parabola 2y = x² and line y = mx + 2. (U may find only the x coordinate).

U will get x = m + (m² + 4)^1/2

Since x₂> x₁,

So, x₂ = m + (m² + 4)^1/2

and x₁ = m - (m² + 4)^1/2

I = (mx²/2) + 2x + (x³/6) limit of x from x₁ to x₂

Put the limits and then find dI/dm and put it equal to 0 to get a value of m.

Check thru double derivative that d²I/dm² > 0.

So, value of m corresponds to minimum value of I.

shashank

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Joined: 26 Dec 2009

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16 Mar 2010 11:11:19 IST

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hey thanks deepak...........

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