JEE MAIN - Mathematics (2002 - No. 23)
The area bounded by the curves $$y = \ln x,y = \ln \left| x \right|,y = \left| {\ln {\mkern 1mu} x} \right|$$ and $$y = \left| {\ln \left| x \right|} \right|$$ is :
$$4$$sq. units
$$6$$sq. units
$$10$$sq. units
none of these
Explanation
First we draw each curve as separate graph
NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.
Clearly the bounded area is as shown in the following figure.
Required area $$ = 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$$
$$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$$ sq. units
NOTE : Graph of $$y = \left| {f\left( x \right)} \right|$$ can be obtained from the graph of the curve $$y = f\left( x \right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis.
Clearly the bounded area is as shown in the following figure.
Required area $$ = 4\int\limits_0^1 {\left( { - \ln x} \right)} dx$$
$$ = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,$$ sq. units
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