Mathematics - Area Under Curves Question with Solution | TestHub

1. $y=\ln x$: This function is defined only for $x>0$. It passes through $(1,0)$.

- As $x\to 0^{+}$, $y\to -\infty$.

- As $x\to \infty$, $y\to \infty$.

2. $y=\ln |x|$: This function is defined for $x\ne 0$.

- For $x>0$, $y=\ln x$.

- For $x<0$, $y=\ln (-x)$. This part of the graph is a reflection of $y=\ln x$ about the y-axis. It passes through $(-1,0)$.

3. $y=|\ln x|$: This function is defined for $x>0$.

- If $\ln x\ge 0$, i.e., $x\ge 1$, then $y=\ln x$.

- If $\ln x<0$, i.e., $0<x<1$, then $y=-\ln x$. This part of the graph is a reflection of $y=\ln x$ about the x-axis for $0<x<1$.

4. $y=|\ln |x||$: This function is defined for $x\ne 0$.

- For $x>0$, $y=|\ln x|$.

- For $x<0$, $y=|\ln (-x)|$. This part of the graph is a reflection of $y=|\ln x|$ about the y-axis.

Let's sketch these graphs.

Graph of $y=\ln x$:

It starts from $-\infty$ at $x=0$, passes through $(1,0)$, and goes to $\infty$.

Graph of $y=\ln |x|$:

It is the graph of $y=\ln x$ for $x>0$ and its reflection about the y-axis for $x<0$. It passes through $(1,0)$ and $(-1,0)$.

Graph of $y=|\ln x|$:

For $x\ge 1$, it is $y=\ln x$. For $0<x<1$, it is $y=-\ln x$. This means the part of $y=\ln x$ below the x-axis (for $0<x<1$) is reflected above the x-axis. So, $y=|\ln x|$ is always non-negative. It passes through $(1,0)$.

Graph of $y=|\ln |x||$:

This is the graph of $y=|\ln x|$ for $x>0$ and its reflection about the y-axis for $x<0$. This function is always non-negative. It passes through $(1,0)$ and $(-1,0)$.

Now let's identify the regions bounded by these curves.

The region is symmetric with respect to the y-axis. We can calculate the area for $x>0$ and then multiply by 2.

For $x>0$:

The curves are $y=\ln x$ and $y=|\ln x|$.

- For $0<x<1$: $\ln x<0$, so $|\ln x|=-\ln x$. In this interval, $y=-\ln x$ is above $y=\ln x$.

- For $x\ge 1$: $\ln x\ge 0$, so $|\ln x|=\ln x$. In this interval, $y=\ln x$ and $y=|\ln x|$ are the same.

The region bounded by $y=\ln x$ and $y=|\ln x|$ for $x>0$ is for $0<x<1$.

The area in this region is given by:

$A_1={\int }_0^1(|\ln x|-\ln x)dx$

Since for $0<x<1$, $\ln x<0$, we have $|\ln x|=-\ln x$.

$A_1={\int }_0^1(-\ln x-\ln x)dx={\int }_0^1(-2\ln x)dx=-2{\int }_0^1\ln xdx$

We know that $\int \ln xdx=x\ln x-x+C$.

So, ${\int }_0^1\ln xdx=[x\ln x-x{]}_0^1$.

We need to evaluate ${\lim }_{x\to 0^{+}}(x\ln x)$. Using L'Hopital's rule:

${\lim }_{x\to 0^{+}}x\ln x={\lim }_{x\to 0^{+}}\frac{\ln x}{1/x}={\lim }_{x\to 0^{+}}\frac{1/x}{-1/x^2}={\lim }_{x\to 0^{+}}(-x)=0$.

So, ${\int }_0^1\ln xdx=(1\ln 1-1)-(0\ln 0-0)=(0-1)-(0-0)=-1$.

Therefore, $A_1=-2(-1)=2$ square units.

This is the area bounded by $y=\ln x$ and $y=|\ln x|$ for $x>0$.

Now consider the curves $y=\ln |x|$ and $y=|\ln |x||$.

For $x>0$, these become $y=\ln x$ and $y=|\ln x|$, which we have already analyzed.

For $x<0$:

The curves are $y=\ln (-x)$ and $y=|\ln (-x)|$.

Let $u=-x$. Then as $x$ goes from $0$ to $-1$, $u$ goes from $0$ to $1$.

The area in this region (for $x<0$) is given by:

$A_2={\int }_{-1}^0(|\ln (-x)|-\ln (-x))dx$

Let $t=-x$, so $dt=-dx$. When $x=-1$, $t=1$. When $x=0$, $t=0$.

$A_2={\int }_1^0(|\ln t|-\ln t)(-dt)={\int }_0^1(|\ln t|-\ln t)dt$

This is the same integral as $A_1$.

So, $A_2=2$ square units.

The total area bounded by all four curves is the sum of the areas in the positive x-region and negative x-region.

Total Area $=A_1+A_2=2+2=4$ square units.

Let's visualize the regions:

- Region 1: Bounded by $y=\ln x$ and $y=|\ln x|$ for $0<x<1$. This is the region between $y=\ln x$ (which is negative) and $y=-\ln x$ (which is positive) in the interval $(0,1)$. The area is $2$.

- Region 2: Bounded by $y=\ln (-x)$ and $y=|\ln (-x)|$ for $-1<x<0$. This is the region between $y=\ln (-x)$ (which is negative) and $y=-\ln (-x)$ (which is positive) in the interval $(-1,0)$. The area is $2$.

The curves $y=\ln x$ and $y=\ln |x|$ are the same for $x>0$.

The curves $y=|\ln x|$ and $y=|\ln |x||$ are the same for $x>0$.

So, for $x>0$, the bounded area is between $y=\ln x$ and $y=|\ln x|$. This occurs for $0<x<1$.

The area is ${\int }_0^1(|\ln x|-\ln x)dx={\int }_0^1(-\ln x-\ln x)dx={\int }_0^1-2\ln xdx=-2[x\ln x-x{]}_0^1=-2[(1\ln 1-1)-{\lim }_{x\to 0^{+}}(x\ln x-x)]=-2[(-1)-(0-0)]=2$.

Due to symmetry, for $x<0$, the area bounded by $y=\ln |x|$ and $y=|\ln |x||$ is also 2.

This area is ${\int }_{-1}^0(|\ln (-x)|-\ln (-x))dx$.

Let $u=-x$, $du=-dx$.

${\int }_1^0(|\ln u|-\ln u)(-du)={\int }_0^1(|\ln u|-\ln u)du=2$.

Total area $=2+2=4$ square units.

The final answer is $\boxed{4sq.units}$.