If I_1 = intlimits_0^1 {{e^{ - x}}} {cos ^2}x | vedclass

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(D) For $x \in (0, 1)$,we have $x^3  -x^2 > -x$. Since the exponential function $f(t) = e^t$ is strictly

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$I_3 > I_2 > I_1$

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(D) For $x \in (0, 1)$,we have $x^3 Multiplying by $-1$,we get $-x^3 > -x^2 > -x$. Since the exponential function $f(t) = e^t$ is strictly increasing,we have $e^{-x^3} > e^{-x^2} > e^{-x}$. Now,consider the integrands: For $I_3$,the integrand is $f_3(x) = e^{-x^3}$. For $I_2$,the integrand is $f_2(x) = e^{-x^2} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x^2} \cos^2 x \le e^{-x^2}$. For $I_1$,the integrand is $f_1(x) = e^{-x} \cos^2 x$. Since $0 \le \cos^2 x \le 1$,we have $e^{-x} \cos^2 x \le e^{-x}$. Comparing the functions on $(0, 1)$: $e^{-x^3} > e^{-x^2} \ge e^{-x^2} \cos^2 x$ implies $I_3 > I_2$. Also,$e^{-x^2} \cos^2 x > e^{-x} \cos^2 x$ because $e^{-x^2} > e^{-x}$ for $x \in (0, 1)$. Thus,$I_3 > I_2 > I_1$.

If $I_1 = \int\limits_0^1 {{e^{ - x}}} {\cos ^2}x\,dx$,$I_2 = \int\limits_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$ and $I_3 = \int\limits_0^1 {{e^{ - {x^3}}}} dx$; then

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