<p>Which bound holds for \(I=\displaystyle\int_0^{1/2}e^{x^2}\,dx\)? [JEE Advanced 2015]</p>
Step-by-Step Solution
Key Concept: On [0,1/2]: e^(x^2) ranges from e^0=1 to e^(1/4). So 1 \cdot (1/2) \leq I \leq e^(1/4) \cdot (1/2).
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<p>On $[0,1/2]$: $x^2\in[0,1/4]$, so $e^{x^2}\in[1, e^{1/4}]$.</p>
<p>Integrating: $1\cdot\frac{1}{2}\le I\le e^{1/4}\cdot\frac{1}{2}$, i.e., $\frac{1}{2}\le I\le\frac{e^{1/4}}{2}$.</p>
<p>Strict: $e^{x^2}>1$ for $x>0$, so $I>\frac{1}{2}$. ✓</p>
<p>$$\boxed{\frac{1}{2}<I<\frac{e^{1/4}}{2}}\approx(0.5, 0.643)$$</p>
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Correct Answer: A