Question
The area bounded by the axes of reference and the normal to $$y = {\log _e}x$$ at the point $$\left( {1,\,0} \right)$$ is :
A.
$$1{\text{ uni}}{{\text{t}}^2}$$
B.
$$2{\text{ uni}}{{\text{t}}^2}$$
C.
$$\frac{1}{2}{\text{ uni}}{{\text{t}}^2}$$
D.
none of these
Answer :
$$\frac{1}{2}{\text{ uni}}{{\text{t}}^2}$$
Solution :
$$y = {\log _e}x\,\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{x}\,\,\,\,\,\,\therefore {\left. {\frac{{dy}}{{dx}}} \right)_{1,\,0}} = 1$$
$$\therefore $$ the equation of the normal at $$\left( {1,\,0} \right)$$ is
$$\eqalign{
& y - 0 = - \frac{1}{1}\left( {x - 1} \right){\text{ or }}x + y = 1 \cr
& \therefore {\text{ area}} = \frac{1}{2}.1.1 = \frac{1}{2} \cr} $$