<p>Evaluate \(\displaystyle\int_0^1\frac{dx}{1+x^2}\)</p>
Step-by-Step Solution
Key Concept: \int dx/(1+x^2) = arctan x + C. Evaluate from 0 to 1.
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<p>$\displaystyle\int_0^1\frac{dx}{1+x^2} = [\arctan x]_0^1 = \arctan(1)-\arctan(0) = \frac{\pi}{4}-0 = \boxed{\frac{\pi}{4}}$</p>
<p><em>Bounds check:</em> On $[0,1]$: $\frac{1}{2}\le\frac{1}{1+x^2}\le 1$, so $\frac{1}{2}\le I\le 1$. Indeed $\pi/4\approx 0.785\in[0.5,1]$.✓</p>
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Correct Answer: C