The area bounded by the curve $y=\ln (x)$ and the lines $y=0, y=\ln (3)$ and $x=0$ is equal to:

Question

Aicharya · Accepted Answer

To find the point of intersection of curves $ \begin{aligned} & 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{aligned} $

$\begin{aligned} \text { Required area } & =\int_0^3 \ln (3) d x-\int_1^3 \ln (x) d x \\ = & {[x \ln (3)]_0^3-\left[x \ln (x)-x_1^3=2\right.}\end{aligned}$

The area bounded by the curve $y=\ln (x)$ and the lines $y=0, y=\ln (3)$ and $x=0$ is equal to:

Solution