<p>Which best bounds \(I=\displaystyle\int_0^1\frac{dx}{\sqrt{x^3+1}}\)? [JEE Advanced 2011]</p>
Step-by-Step Solution
Key Concept: On [0,1]: 1 \leq x^3+1 \leq 2, so 1/\sqrt{2} \leq 1/\sqrt{x^3+1} \leq 1. Integrating: 1/\sqrt{2} \leq I \leq 1. More precisely I > 2/3 by estimation.
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<p>On $[0,1]$: $1\le x^3+1\le 2\Rightarrow\frac{1}{\sqrt{2}}\le\frac{1}{\sqrt{x^3+1}}\le 1$.</p>
<p>So $\frac{1}{\sqrt{2}}\le I\le 1$, i.e., $0.707\le I\le 1$.</p>
<p>Better lower bound: On $[0,1]$, $x^3+1\le x+1$ (since $x^3\le x$ for $x\in[0,1]$), so $\frac{1}{\sqrt{x^3+1}}\ge\frac{1}{\sqrt{x+1}}$.</p>
<p>$\int_0^1\frac{dx}{\sqrt{x+1}}=[2\sqrt{x+1}]_0^1=2\sqrt{2}-2\approx 0.828>2/3$. ✓</p>
<p>So $\boxed{2/3 < I < 1}$.</p>
Correct Answer: A