\( \int_{0}^{2} 3 x+\sqrt{4-x^{2}} \) by interpreting the integral in terms of areas.

To solve the integral \( \int_{0}^{2} (3x + \sqrt{4 - x^2}) \, dx \) by interpreting it in terms of areas, we can break it into two parts. The term \( 3x \) represents the area under the line from \( 0 \) to \( 2 \), which forms a triangle. The base is \( 2 \) and the height at \( x = 2 \) is \( 3 \times 2 = 6 \), giving an area of \( \frac{1}{2} \times 2 \times 6 = 6 \). Next, the term \( \sqrt{4 - x^2} \) represents the upper half of a circle with radius \( 2 \). The area of a circle is \( \pi r^2 \), so the area of the upper half is \( \frac{1}{2} \pi (2^2) = 2\pi \). Thus, the total area represented by the integral from \( 0 \) to \( 2 \) is the area of the triangle plus the area of the semicircle: \[ \text{Total Area} = 6 + 2\pi. \] Hence, the value of the integral is: \[ \int_{0}^{2} (3x + \sqrt{4 - x^2}) \, dx = 6 + 2\pi. \]