Question

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

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

Answer: (C)

1

Given curve is $y={\log }_e(x+e)$

Since, the region is bounded by coordinate axes, so the curve touches the coordinate axes.

Thus, we have $x=0$ at some point $y$ and $y=0$ at some point

So, $x=0$ is one of the limit for the bounded area

$y=0⟹{\log }_e(x+e)=0⟹x+e=e^0=1⟹x=1-e$ is the another limit

Since, $1-e<0⟹x=0$ is upper limit and $x=1-e$ is lower limit

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

Put $x+e=t⟹dx=dt$

At $x=1-e,t=1$ and $t=e$ when $x=0$
$\therefore A={\int }_1^e\log tdt=[t\log t-t{]}_1^e$

$=e\log e-e-(\log 1-1)$
$=1$ sq unit