test book

1. chap 1

Your answer is incorrect.

To find the point of intersection,

            log x = (log x)2

\Rightarrow         log x (log x – 1) = 0

\Rightarrow         log x = 0        or    log x = 1

\Rightarrow         x = 1, e



Hence, the required area will be \int_{1}^{e}\int_{\(log x)^2}^{\log x}dydx

= \int_{1}^{e} (\log x - \left ( \log x \right )^2) dx

= \int_{1}^{e} \log x dx - \int_{1}^{e} \left (\log x \right )^2 dx

= \left [ x\log x - x \right ]_{1}^{e} - \left [ x(\log x)^2 - 2x \log x + 2x \right ]_{1}^{e}

= [e - e - (-1)] - [e - 2e + 2e - 2]

= 1 - (e - 2)

= (3 - e) sq. units