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  2. Stuck here, help me understand: - Integral Calculus - JEE Main-4
# Engineering

If I_{1}=\int_{0}^{1}e^{-1}\cos ^{2}xdx,

   I_{2}=\int_{0}^{1}e^{-x^{2}}\cos ^{2}xdx  and

I_{3}=\int_{0}^{1}e^{-x^{3}}dx ; then:

 

  • Option 1)

    I_{2}> I_{3}> I_{1}

  • Option 2)

    I_{2}> I_{1}> I_{3}

  • Option 3)

    I_{3}> I_{2}> I_{1}

  • Option 4)

    I_{3}> I_{1}> I_{2}

 

Answers (2)
A Anuj Pandey
H Himanshu

As we learned,

 

Properties of Definite Integration -

If f(x)\geqslant g(x) for

all x\in \left [ a,b \right ] then

\int_{a}^{b}f\left ( x \right )dx\geqslant \int_{a}^{b}g(x)dx

-

 

 

 In the interval [0,1]

e^{x}> e^{x^{2}}> e^{x^{3}}

or  e^{-x}< e^{-x^{2}}< e^{-x^{3}}

Thus e^{-x}\cos ^{2}x< e^{-x^{2}}\cos ^{2}x< e^{-x^{3}}

On Integration,

I_{1}< I_{2}< I_{3}

 


Option 1)

I_{2}> I_{3}> I_{1}

Option 2)

I_{2}> I_{1}> I_{3}

Option 3)

I_{3}> I_{2}> I_{1}

Option 4)

I_{3}> I_{1}> I_{2}

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