If (a,b) be the orthocentre of the triangle whose vertices are (1,2),(2,3) and (3,1), and I_1=∫_a^b xsin(4x-x^2 )dx,□( ) I_2=∫_a^b sin(4x-x^2 )dx , then 36 I_1/I_2 is equal to : (1) 72 (2) 88 (3) 80 (4) 66 Ans. (1) Sol. Equation of CE y-1=-(x-3) x+y=4 orthocentre lies on the line x+y=4 so, a+b=4 I_1=∫_a^b xsin(x(4-x))dx Using king rule I_1=∫_a^b (4-x)sin(x(4-x))dx (i) + (ii) 2I_1=∫_a^b 4sin(x(4-x))dx 2I_1=4I_2 I_1=2I_2 I_1/I_2 =2 (36I_1)/I_2 =72
