Consider the integrals $I_1 = \int_0^1 e^{-x}\cos^2 x dx, I_2 = \int_0^1 e^{-x^2} \cos^2 x dx, I_3 = \int_0^1 e^{-x} dx$ and $I_4 = \int_0^1 e^{-2/2} dx$. Then:
Step-by-Step Solution
Key Concept: Use the bounds $e^{-x^2} > e^{-x}$ for $x \in (0,1)$ and $\cos^2 x < 1$ to establish ordering between integrals.
We compare these integrals by analyzing bounds on $[0,1]$. For $I_1$ vs $I_3$: since $\cos^2 x \leq 1$, we have $I_1 = \int_0^1 e^{-x}\cos^2 x dx \leq \int_0^1 e^{-x} dx = I_3$. For $I_2$ vs $I_1$: on $[0,1]$, $e^{-x^2} > e^{-x}$ (since $-x^2 > -x$ when $0 < x < 1$), so $I_2 > I_1$. For $I_4$ vs $I_2$: interpreting $I_4 = \int_0^1 e^{-x^2} dx$, we have $I_4 = \int_0^1 e^{-x^2} dx > \int_0^1 e^{-x^2}\cos^2 x dx = I_2$ since $\cos^2 x < 1$ on $(0,1)$. Therefore $I_4 > I_2 > I_1$ and $I_3 > I_1$, plus $I_3 = 1-e^{-1} \approx 0.632 > e^{-1} \approx 0.368$.
Correct Answer: 1,2,3