Question

Let $${I_1} = \int_0^1 {{e^{ - {x^2}}}} dx,\,{I_2} = \int_0^1 {{e^{ - x}}} {\cos ^2}x\,dx$$        and $${I_3} = \int_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,} dx.$$     Then :

A. $${I_1} < {I_2} < {I_3}$$
B. $${I_3} < {I_2} < {I_1}$$
C. $${I_2} < {I_1} < {I_3}$$
D. $${I_2} < {I_3} < {I_1}$$  
Answer :   $${I_2} < {I_3} < {I_1}$$
Solution :
$$\eqalign{ & {\text{In }}\left( {0,\,1} \right),\,{e^{ - {x^2}}} > {e^{ - {x^2}}}{\cos ^2}x > 0\,\,\,\,\,\,\,\,\therefore {I_1} > {I_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{e^{ - {x^2}}} > {e^{ - x}} > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore {e^{ - {x^2}}}{\cos ^2}x > {e^{ - x}}{\cos ^2}x > 0 \cr & \therefore {I_3} > {I_2} \cr & \therefore {I_2} < {I_3} < {I_1} \cr} $$

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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