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If $f\left ( x \right ) = 2 \:\ln \left ( \sqrt{e^{x}} \right )$, what is the area bounded by $f\left ( x \right )$ for the interval $\left [ 0,2 \right ]$ on the $x$ – axis?

  1. $\frac{1}{2}$
  2. $1$
  3. $2$
  4. $4$
in Quantitative Aptitude 27.4k points
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1 Answer

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Given that, $f\left ( x \right ) = 2 \:\ln \left ( \sqrt{e^{x}} \right );\; x \in [0,2]$

$\Rightarrow f(x) = 2 \ln(e^{x})^{\frac{1}{2}}$

$\Rightarrow f(x) = 2 \times \frac{1}{2} \ln e^{x}$

$\Rightarrow f(x) =  \log_{e} e^{x}\quad [{\color{Green}{\because \ln x = \log_{e}x}}]$

$\Rightarrow f(x) =  x\log_{e} e \quad [{\color{Purple}{\because \log_{b}a^{x} = x\log_{b}a}}]$

$\Rightarrow {\color{Blue}{\boxed{f(x) = x}}} \quad [{\color{Teal}{\because \log_{b}b = 1}}]$

We can draw the diagram,

Now we can define the area bounded by $f(x)$ for the interval $[0,2]$ on the $x$-axis.

Required area $ = \displaystyle{} \int_{0}^{2} f(x) dx  = \int_{0}^{2} x dx = \left[ \dfrac{x^{2}}{2}\right]_{0}^{2} = \left[ \dfrac{2^{2}}{2} \; –\; \dfrac{0^{2}}{2}\right] = 2 \;\text{unit}^{2}.$

Correct Answer $:\text{C}$

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