Question

The area (in sq units) bounded by the curve $y=1+{\log }_{e}x$, the $X$-axis and the straight line$x=e$ is equal to

• A. $3e–2$
• B. $e$
• C. $e-\frac{1}{e}$
• D. $e+\frac{1}{e}$

Answer 1

The correct option is D

$e+\frac{1}{e}$

Explanation for Correct Answer:

Step 1:Determine the point of intersection of the curves.

Given, that curves are

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

$x=e\to \left(ii\right)$

To determine the point of intersection of the curves,

Substitute equation $x=e$ in equation $\left(i\right)$

$$\begin{gathered} y=1+{\log }_{e}x \\ =1+{\log }_{e}e \\ =1+1\left[\because {\log }_{e}e=1\right] \\ =2 \end{gathered}$$

$\therefore y=2$

Substituting equation $y=0$ in equation $\left(i\right)$

$$\begin{gathered} y=1+{\log }_{e}x \\ \Rightarrow 0=1+{\log }_{e}x \\ \Rightarrow -1={\log }_{e}x \\ \Rightarrow e^{-1}=e^{{\log }_{e}x}\left[\because \text{Rasing}e\text{power on both sides}\right] \\ \Rightarrow x=e^{-1}\left[\because e^{{\log }_{e}x}=x\right] \end{gathered}$$

The graph of the given curves, is as shown:

Step 2:Determining the area of the bounded region.

Now, the area of the bounded region is:

$$\begin{gathered} A={\int }_{e^{-1}}^{e}ydx \\ ={\int }_{e^{-1}}^e\left(1+{\log }_{e}x\right)dx \\ ={\left[x+x{\log }_{e}x-x\right]}_{e^{-1}}^e \\ ={\left[x{\log }_{e}x\right]}_{e^{-1}}^e \\ =\left[e{\log }_{e}e\right]-\left[e^{-1}{\log }_e{e}^{-1}\right] \\ =e\times 1-\left[e^{-1}\left(-1\right){\log }_e{e}^1\right]\left[\because {\log }_{e}e=1;{\log }_e{e}^x=x{\log }_{e}e\right] \\ =e-e^{-1}\left(-1\right) \\ =e+\frac{1}{e} \end{gathered}$$

Therefore the area bounded by the given curve is equal to $e+\frac{1}{e}\text{square units}$

Hence, the correct answer is option (D).