Question

If ${{I}_{1}}=\int\limits_{0}^{1}{{{e}^{-x}}{{\cos }^{2}}xdx}$ , ${{I}_{2}}=\int\limits_{0}^{1}{{{e}^{-{{x}^{2}}}}{{\cos }^{2}}xdx}$ and ${{I}_{3}}=\int\limits_{0}^{1}{{{e}^{-{{x}^{3}}}}dx}$, then
( a )${{I}_{2}}>{{I}_{3}}>{{I}_{1}}$
( b ) ${{I}_{3}}>{{I}_{1}}>{{I}_{2}}$
( c ) ${{I}_{2}}>{{I}_{1}}>{{I}_{3}}$
( d ) ${{I}_{3}}>{{I}_{2}}>{{I}_{1}}$

Answer 1

Hint: We can solve this question by simple comparing functions whose integration is to be done by simple rules of algebra and real numbers. We will compare function for open interval ( 0, 1 ) and then hence, we will check which function greater to which function.

Complete step by step answer:
In all three integration, limits is from 0 to 1, which means integration of function from limits 0 to 1 gives the, area under curves between interval 0 to 1,
Now, as limits are equal so, we can solve this question by comparing the functions without directly solving them.
Now, we know that if x belongs to interval (0,1),
Then, $x>{{x}^{2}}$ , for $x\in (0,1)$
Or, $-x<-{{x}^{2}}$ , for $x\in (0,1)$
Or, ${{e}^{-x}}<{{e}^{-{{x}^{2}}}}$, for $x\in (0,1)$.
As, $\cos x$ is even function as $\cos x=\cos (-x)$ , in its domain,
So, ${{e}^{-x}}\cos x<{{e}^{-{{x}^{2}}}}\cos x$, for $x\in (0,1)$
So, as limits of integration are same for both ${{I}_{1}}$ and ${{I}_{2}}$, so we can say that ${{I}_{2}}>{{I}_{1}}$….. ( i )
Now, again for x belonging to ( 0, 1 )
We can say that,
$x>{{x}^{3}}$ , for $x\in (0,1)$
Or, $-x<-{{x}^{3}}$ , for $x\in (0,1)$
Or, ${{e}^{-x}}<{{e}^{-{{x}^{3}}}}$, for $x\in (0,1)$.
As, $\cos x$, for $x\in (0,1)$, $\cos x$ is decreasing function.
So, ${{I}_{3}}>{{I}_{1}}$……( ii )
So, ${{e}^{-x}}{{\cos }^{2}}x<{{e}^{-{{x}^{3}}}}$, for $x\in (0,1)$.
And for , ${{e}^{-x}}^{^{2}}<{{e}^{-{{x}^{3}}}}$, for $x\in (0,1)$.
We can say that, ${{e}^{-x}}^{^{2}}{{\cos }^{2}}x<{{e}^{-{{x}^{3}}}}$, for $x\in (0,1)$.
So, ${{I}_{3}}>{{I}_{2}}$……. ( iii )
Thus, from equation ( I ), ( ii ) and ( iii ), we get
${{I}_{3}}>{{I}_{2}}>{{I}_{1}}$

So, the correct answer is “Option D”.

Note: Above question can be solved by another rule of integration called product rule of integration which states that $\int{u\cdot vdx=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}}}dx$ , where u and v are function of x and where priority of function u ( x ) decreases from ILATE, where I = inverse function, L = logarithmic function, A = algebraic function and E = exponential function.