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Consider the integrals I_{1}=\int_{0}^{1}e^{-x}\cos ^{2}x\, dx,I_{2}=\int_{0}^{1}\cos ^{2}x\, dx,I_{3}=\int_{0}^{1}e^{-\frac{x^{2}}{2}}\cos ^{2}x\, dx,I_{4}=\int_{0}^{1}e^{-\frac{x^{2}}{2}}\: dx Then

 

Option: 1

I_{2}>I_{4}>I_{1}>I_{3}
 


Option: 2

I_{2}<I_{4}<I_{1}<I_{3}


Option: 3

I_{1}<I_{2}<I_{3}<I_{4}

 


Option: 4

I_{1}>I_{2}>I_{3}>I_{4}


Answers (1)

best_answer

 

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

-

 

 

Since 0<x<1\; \; \; \; \; \; \Rightarrow x>x^{2}>\frac{x^{2}}{2}

\Rightarrow -x<-x^{2}<-\frac{x^{2}}{2}\; \; \; \; \Rightarrow e^{-x} < e^{-x^{2}}<e^{\frac{x^{2}}{2}}

\Rightarrow e^{-x}\cos ^{2}x<e^{-x^{2}}\cos ^{2}x<e^{\frac{x^{2}}{2}}\cos ^{2}x<e^{\frac{x^{2}}{2}}

\therefore \int_{0}^{1}e^{-x}\cos ^{2}x\; dx<\int_{0}^{1}e^{-x^{2}}\cos ^{2}x\; dx

<\int_{0}^{1}e^{-\frac{x^{2}}{2}}\cos ^{2}x\; dx <\int_{0}^{1}e^{-\frac{x^{2}}{2}}dx

\therefore I_{1}<I_{2}<I_{3}<I_{4}

Posted by

Rakesh

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