Answer
$$A = 1$$
Work Step by Step
$$\eqalign{
& y = - \ln x,{\text{ }}0 < x \leqslant 1 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left( { - \ln x} \right)} dx \cr
& A = - \int_0^1 {\ln x} dx \cr
& \ln x{\text{ has an infinite discontinuity at }}x = 0,{\text{so we can write}} \cr
& A = - \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\ln x} dx \cr
& {\text{Integrate}} \cr
& A = - \mathop {\lim }\limits_{a \to {0^ + }} \left[ {x\ln x - x} \right]_a^1 \cr
& A = - \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\left( {1\ln 1 - 1} \right) - \left( {a\ln a - a} \right)} \right] \cr
& A = - \mathop {\lim }\limits_{a \to {0^ + }} \left[ { - 1 - \left( {a\ln a - a} \right)} \right] \cr
& {\text{Evaluate }}a \to {0^ + } \cr
& A = - \left[ { - 1 - \left( {\underbrace {{0^ + }\ln {0^ + }}_0 - {0^0}} \right)} \right] \cr
& A = 1 \cr} $$
