Question

The area of the figure bounded by the curves $y=\ln x\And y={{\left( \ln x \right)}^{2}}$ is;
A) e + 1
B) e – 1
C) 3 – e
D) 1

Answer 1

In the given question, we have been asked to find the area bounded by the given curve. In order to find the area first we need to find the point of intersection of two given functions i.e. $y=\ln x\ and\ y={{\left( \ln x \right)}^{2}}$. Then we can calculate the area between two curves by the difference between the definite integral of the given two functions.

Complete step by step solution:
We have given that,
$y=\ln x\ and\ y={{\left( \ln x \right)}^{2}}$
The graph of the function$y=\ln x\ and\ y={{\left( \ln x \right)}^{2}}$is as follows;

We can observe that,
The point of intersection of the given curves that is $y=\ln x\ and\ y={{\left( \ln x \right)}^{2}}$are (1, 0) and (e, 1).
Now,
The required area is;
$area=\int\limits_{1}^{e}{\left( \ln x-{{\left( \ln x \right)}^{2}} \right)dx}$
Integrating the above integral by using by-parts method,
Formula of integration by parts as follows;
$\int{f\left( x \right)g'\left( x \right)dx=f\left( x \right)g\left( x \right)-\int{f'\left( x \right)g\left( x \right)dx}}$
Thus, integrating the resultant expression, we obtain
$\Rightarrow A=\left( x\ln x-x \right)_{1}^{e}-\left( x{{\left( \ln x \right)}^{2}}-\int\limits_{1}^{e}{2\ln xdx} \right)$
$\Rightarrow A=\left( x\ln x-x \right)_{1}^{e}-\left( x{{\left( \ln x \right)}^{2}}-2x\ln x+2x \right)_{1}^{e}$
Therefore,
$\Rightarrow A=\left( x\ln x-x-x{{\left( \ln x \right)}^{2}}+2x\ln x-2x \right)_{1}^{e}$
Considering the limits, we get
$\Rightarrow A=\left| 3x\left( \ln x-1 \right) \right|_{1}^{e}-\left| x{{\left( \ln x \right)}^{2}} \right|_{1}^{e}=3-e$
Therefore,

The area bounded by the region $y=\ln x\ and\ y={{\left( \ln x \right)}^{2}}$ is equal to 3 – e.
Hence, this is the required answer.

Note:
Integration is used to find the area bounded by the given curves. In order to find the area between the two curves is the difference between the area below the upper curve and the area below the lower curve. In mathematical form, we are computing the area as follows;
If y = f (x) and y = g (x) i.e. the two given function, ${{x}_{1}}$ and ${{x}_{2}}$are the two given limits.
Then area between the two curves is as follows;
Area between the two curves = $\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( f\left( x \right)-g\left( x \right) \right)}dx$.