Mathematics - Area Under Curves Question with Solution | TestHub

To find the point of intersection of curves $$\begin{array}{rl}& y=\ln (x)\text{ and }y=\ln (3),\text{ put }\ln (x)=\ln (3)\\ \Rightarrow & \ln (x)-\ln (3)=0\\ \Rightarrow & \ln (x)-\ln (3)=\ln (1)\\ \Rightarrow & \frac{x}{3}=1,\Rightarrow x=3\end{array}$$
$\begin{array}{rl}\text{ Required area }& ={\int }_0^3\ln (3)dx-{\int }_1^3\ln (x)dx\\ =& [x\ln (3){]}_0^3-[x\ln (x)-x_1^3=2\end{array}$