Question

The areabounded by the axes of reference and the normal to y=logxat the point (1, 0) is

Answer 1

curve  y = log x    we take the base  as  e  (not decimal)
slope of tangent =  dy/dx  = 1 / x
           at (1,  0)  slope =  1

so slope of normal at  (1,0) =  -1  as the product of slopes is -1..

equation of normal is  y = - x + c
   as it passes throug h  (1,0),  c = 1
         so normal is :  y = 1 - x
     y-intercept  of this  normal is :  1

Area of the traingle formed by the normal to the curve, and the axes is
       1/2 * 1 * 1 = 1/2