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 x1

value of  m is minimum-?

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.

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