Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

1

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

Put $x+e=t\Rightarrow dx=dt$
so, $x\to 1-e\Rightarrow t\to 1$ and $x\to 0\Rightarrow t\to e$

Therefore,

area $={\int }_1^e\log tdt$

$={[t\log t-t]}_1^e$

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