MCQ
The area enclosed between the curve $\text{y}=\log_{\text{e}}(\text{x}+\text{e}),\text{x}=\log_\text{e}\Big(\frac{1}{\text{y}}\Big)$ and the $x-$axis is:
- ✓$2$
- B$1$
- C$4$
- Dnone of these
✓
Answer
Correct option: A.
$2$
The point of intersection of the curves $\text{y}=\log_{\text{e}}(\text{x}+\text{e})$ and $\text{x}=\log_\text{e}\Big(\frac{1}{\text{y}}\Big) $
$\text{y}=\log_{\text{e}}(\text{x}+\text{e})$
$\Rightarrow\text{x}+\text{e}=\text{e}^{\text{y}}$
$\Rightarrow\text{x}=\text{e}^{\text{y}}-\text{e}$
and $\text{x}_{2} = \log_\text{e}\Big(\frac{1}{\text{y}}\Big)$
Therefore, area of the required region,
$\text{A} = \int\limits^1_0(\text{x}_2-\text{x}_1)\text{ dy} \Big[$where$, = \text{x}_1 = \text{e}^{\text{y}}-\text{e}$ and $\text{x}_2= \log_\text{e}\Big(\frac{1}{\text{y}}\Big)\Big]$
$\text{A} = \int\limits^1_0\log_\text{e}\Big(\frac{1}{\text{y}}\Big)\text{ dy}-\int\limits^1_0(\text{e}^\text{y}-\text{e})\text{ dy}$
$\text{A}=\int\limits^1_0\log_\text{e}\Big(\frac{1}{\text{y}}\Big)\text{ dy}-\big[\text{e}^{\text{y}}-\text{ey}\big]\ ...(\text{i})$
Let $\text{I} = \int\log_\text{e}\Big(\frac{1}{\text{y}}\Big)\text{dy}$
Putting $\frac{1}{\text{y}}=\text{t}$
Therefore, integral becomes
$\text{I} = \int-\frac{1}{\text{t}^{2}}\log_\text{e}\text{t}\text{ dt} $
$= -\log_\text{e}\text{t}\int\frac{1}{\text{t}^{2}}\text{ dt}-\int\frac{1}{\text{t}}\times\frac{1}{\text{t}}\text{ dt}$
$= \frac{1}{\text{t}}\log_\text{e}\text{t}+\frac{1}{\text{t}}$
$= \text{y}\log_\text{e}\frac{1}{\text{y}}+\text{y}$
Now$, (i)$ becomes
$\text{A} = \Big[\text{y}\log_\text{e}\frac{1}{\text{y}}+\text{y}\Big]^1_0-\big[\text{e}^\text{y}-\text{ey}\big]^1_0$
$=\Big[\text{y}\log_\text{e}\Big(\frac{1}{\text{y}}\Big)+\text{y}-\text{e}^{\text{y}}+\text{ey}\Big]^1_0$
$=\big[\log_\text{e}(1)+1-\text{e}^1+\text{e}(1)\big]-\big[0+0-\text{e}^{(0)}+\text{e}(0)\big]$
$= 2$
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