The curve y=a x^2+b x passes through the point (1,2) and lies above the X -axis for 0 x 8 . If the area enclosed by this curve, the X -axis and the line x=6 is 108 square units, then 2 b-a=

Step-by-step solution

Given, curve y=a x^2+b x passes through point (1,2) . 2=a(1)^2+b(1) aligned & Given, area under curve = ₀^6 (a x^2+b x ) d x=108 & [ a x^3 3 + b x^2 2 ]₀^6=108 & 72 a+18 b=108 aligned By solving Eqs. (i) and (ii), we get aligned & a= 4 3 and b= 2 3 & 2 b-a=2 2 3 - 4 3 = 4 3 - 4 3 =0 aligned