<p>Evaluate \(\displaystyle\int_0^1\sqrt{1-x^2}\,dx\)</p>
Step-by-Step Solution
Key Concept: y = \sqrt{1-x^2} is the upper semicircle of radius 1. \int_0^1 = area of quarter circle = \pi \cdot 1^2/4 = \pi/4.
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<p><strong>Geometric method:</strong> $y=\sqrt{1-x^2}$ is the upper semicircle $x^2+y^2=1$, $y\ge 0$.</p>
<p>$\int_0^1\sqrt{1-x^2}\,dx$ = area of the quarter circle in the first quadrant = $\dfrac{\pi(1)^2}{4} = \boxed{\dfrac{\pi}{4}}$</p>
<p><strong>Algebraic:</strong> Let $x=\sin\theta$, $dx=\cos\theta\,d\theta$:</p>
<p>$$= \int_0^{\pi/2}\cos^2\theta\,d\theta = \frac{1}{2}\int_0^{\pi/2}(1+\cos 2\theta)\,d\theta = \frac{1}{2}\left[\theta+\frac{\sin 2\theta}{2}\right]_0^{\pi/2} = \frac{\pi}{4}$$</p>
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Correct Answer: D