<p>Which of the following is the tightest correct bound for \(I=\displaystyle\int_0^1 e^{-x^2}\,dx\)?</p>
Step-by-Step Solution
Key Concept: On [0,1]: e^(-1) \leq e^(-x^2) \leq 1. Integrating: e^(-1) \leq I \leq 1. More precisely I \in (1-1/e, 1).
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<p>On $[0,1]$: $0\le x^2\le 1\Rightarrow e^{-1}\le e^{-x^2}\le 1$.</p>
<p>Integrating: $e^{-1}\le I\le 1$, i.e. $\frac{1}{e}\le I\le 1$.</p>
<p>Strict inequalities hold since $e^{-x^2}=e^{-1}$ only at $x=1$ (measure zero) and $e^{-x^2}=1$ only at $x=0$.</p>
<p>Also $I = 1-\int_0^1(1-e^{-x^2})dx$. Since $1-e^{-x^2}>0$ for $x>0$, we get $I<1$.</p>
<p>Lower bound refinement: $I > \int_0^1(1-x^2)dx = 1-\frac{1}{3}=\frac{2}{3} > 1-\frac{1}{e}\approx 0.632$. ✓</p>
<p>So $\boxed{1-\frac{1}{e} < I < 1}$. ✓</p>
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Correct Answer: ['A']