\( A = \int _ { 0 } ^ { 2 } \sqrt { 4 - x ^ { 2 } } d x \)
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To solve the integral \( A = \int _ { 0 } ^ { 2 } \sqrt { 4 - x ^ { 2 } } d x \), we can recognize that the integrand represents the upper half of a circle with radius 2, centered at the origin. Thus, the area under the curve from \( x = 0 \) to \( x = 2 \) corresponds to a quarter of the full circle's area. The area of the full circle is \( \pi r^2 = \pi (2^2) = 4\pi \). Therefore, the area of the quarter circle is \( \frac{1}{4} (4\pi) = \pi \). So, \( A = \pi \). To verify our calculations or for further understanding, you can also evaluate the integral using trigonometric substitution, specifically letting \( x = 2\sin(\theta) \). This substitution transforms the limits and the integrand, leading to a trigonometric integral that can be solved. This approach can solidify your grasp of integral techniques, especially in cases involving geometric interpretations!