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

We start with the integral \[ \int_{0}^{2} \left(3x+\sqrt{4-x^2}\right) dx, \] which we split as \[ \int_{0}^{2} 3x\, dx + \int_{0}^{2} \sqrt{4-x^2}\, dx. \] **Step 1. Evaluate \(\int_{0}^{2} 3x\, dx\):** The integral of \(3x\) is \[ \int 3x\, dx = \frac{3}{2}x^2. \] Evaluating from \(x=0\) to \(x=2\): \[ \frac{3}{2}(2^2) - \frac{3}{2}(0^2) = \frac{3}{2}\times 4 = 6. \] **Step 2. Evaluate \(\int_{0}^{2} \sqrt{4-x^2}\, dx\):** The expression \(\sqrt{4-x^2}\) represents the upper half of a circle with radius \(2\). The full circle has area \[ \pi (2)^2 = 4\pi. \] Since the graph of \(\sqrt{4-x^2}\) gives half of the circle (only the top half) and we integrate from \(x=-2\) to \(x=2\) to cover the entire semicircle, the area of the semicircle is \[ \frac{1}{2}(4\pi) = 2\pi. \] However, the integral is from \(x=0\) to \(x=2\), which corresponds to one quarter of the full circle. Therefore, its area is \[ \frac{1}{4}(4\pi) = \pi. \] **Step 3. Combine the results:** Adding both parts, we have \[ 6 + \pi. \]