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)

$1$

Answer 2

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

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