Question

You are given a curve, $y=ln(x+e)$ . What will be the area enclosed between this curve and the coordinate axes?

• A. $1$
• B. $0$
• C. $2e$
• D. $e-1$

Answer 1

Answer: (A)

Given curve, $y=$ ln $(x+e)$
Curve cuts $x$-axis at $(1-e,0)$ and y-axis at $(0,1)$
$\therefore$ Required area $=\underset{1-e}{\overset{0}{\int }}1\cdot ln\left(x+e\right)dx$
$={\left[ln\left(x+e\right)\cdot x\right]}_{1-e}^0-{\int }_{1-e}^0\frac{1}{x+e}\cdot xdx$
$=0-\underset{1-e}{\overset{0}{\int }}\left(\frac{x+e}{x+e}-\frac{e}{x+e}\right)dx$
$=-\underset{1-e}{\overset{0}{\int }}1dx+\underset{1-e}{\overset{0}{\int }}\frac{e}{x+e}dx$

$={\left[-x\right]}_{1-e}^0+{\left[elog\left(x+e\right)\right]}_{1-e}^0$
$=0+1-e+eloge-elog\left(1-e+e\right)$
$=1-e+e=1$