Question

The area bounded by the curves $$y = \ln \,x,\,y = \ln \left| x \right|,\,y = \left| {\ln \,x} \right|$$      and $$y = \left| {\ln \left| x \right|} \right|$$   is-

A. $$4$$ square units  
B. $$6$$ square units
C. $$10$$  square units
D. none of these
Answer :   $$4$$ square units
Solution :
First we draw each curve as separate graph
Application of Integration mcq solution image
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.
Application of Integration mcq solution image
$$\eqalign{ & {\text{Required area}}\, = 4\int\limits_0^1 {\left( { - \ell nx} \right)dx} \cr & = - 4\left[ {x\,\ell n\,x - x} \right]_0^1 \cr & = 4\,{\text{sq}}{\text{. units}} \cr} $$

Releted MCQ Question on
Calculus >> Application of Integration

Releted Question 1

The area bounded by the curves $$y = f\left( x \right),$$   the $$x$$-axis and the ordinates $$x = 1$$  and $$x = b$$  is $$\left( {b - 1} \right)\sin \left( {3b + 4} \right).$$     Then $$f\left( x \right)$$  is-

A. $$\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
B. $$\sin \,\left( {3x + 4} \right)$$
C. $$\sin \,\left( {3x + 4} \right) + 3\left( {x - 1} \right)\cos \left( {3x + 4} \right)$$
D. none of these
Releted Question 2

The area bounded by the curves $$y = \left| x \right| - 1$$   and $$y = - \left| x \right| + 1$$   is-

A. $$1$$
B. $$2$$
C. $$2\sqrt 2 $$
D. $$4$$
Releted Question 3

The area bounded by the curves $$y = \sqrt x ,\,2y + 3 = x$$    and $$x$$-axis in the 1st quadrant is-

A. $$9$$
B. $$\frac{{27}}{4}$$
C. $$36$$
D. $$18$$
Releted Question 4

The area enclosed between the curves $$y = a{x^2}$$   and $$x = a{y^2}\left( {a > 0} \right)$$    is 1 sq. unit, then the value of $$a$$ is-

A. $$\frac{1}{{\sqrt 3 }}$$
B. $$\frac{1}{2}$$
C. $$1$$
D. $$\frac{1}{3}$$

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