Question

The area enclosed between the curve $y={\log }_e\left(x+e\right)$ and the coordinate axes is

• A. $4\text{sq units}$
• B. $3\text{sq units}$
• C. $2\text{sq units}$
• D. $1\text{sq units}$

Answer 1

The correct option is D

$1\text{sq units}$

Explanation for the correct option:

Find the area enclosed between the curve .

Step 1: Find the points where the curve intersects the coordinate axes.

Consider the given Equation as,

$y={\log }_e\left(x+e\right)\to \left(1\right)$

This curve touches the coordinate axis.

For some value of $y$, let $x=0$, we get $y={\log }_e(0+e)={\log }_{e}e=1$

Hence one limit point is $x=0,y=1$

Now Put $y=0$ in Equation (1), we get

$$\begin{gathered} \Rightarrow 0={\log }_e\left(x+e\right) \\ \Rightarrow x+e=e^0 \\ \Rightarrow x+e=1 \\ \Rightarrow x=1-e\left[1-e<0\right] \end{gathered}$$

So, the is another limit point is $x=1-e,y=0$

The graph of the curve is shown as

Step 2: Find the area of shaded portion

Now the area of a shaded portion is,

$A={\int }_{1-e}^0{\log }_e\left(x+e\right)dx$

Put $x+e=t$

Then, $\Rightarrow dx=dt$

Now at $x=1-e,t=1$ and at $x=0,t=e$

So, the area is

$$\begin{gathered} A={\int }_1^e{\log }_{e}tdt \\ A={\left[t{\log }_{e}t-t\right]}_1^e \\ A=\left(e{\log }_{e}e-e\right)-\left(1{\log }_{e}1-1\right) \\ A=\left(e\times 1-e\right)-\left(1\times 0-1\right)\left[\because {\log }_{e}e=1;{\log }_{e}1=0\right] \\ a=0-\left(-1\right) \\ A=1\text{sq unit} \end{gathered}$$

Hence, option (D) is the correct answer.