Evaluate the integral by interpreting it in terms of areas. \[ \int_{-7}^{0}\left(3+\sqrt{49-x^{2}}\right) d x \]

To evaluate the integral \[ \int_{-7}^{0}\left(3+\sqrt{49-x^{2}}\right) \, dx, \] we can interpret it in terms of geometric areas. 1. **Separate the Integral:** The integral can be split into two parts: \[ \int_{-7}^{0} 3 \, dx + \int_{-7}^{0} \sqrt{49 - x^2} \, dx. \] 2. **Evaluate the First Integral:** \[ \int_{-7}^{0} 3 \, dx = 3 \times (0 - (-7)) = 3 \times 7 = 21. \] This represents the area of a rectangle with height 3 and width 7. 3. **Evaluate the Second Integral:** \[ \int_{-7}^{0} \sqrt{49 - x^2} \, dx. \] The function \( \sqrt{49 - x^2} \) describes the upper semicircle of a circle with radius 7 centered at the origin. The integral from \(-7\) to \(0\) represents a quarter of the area of the full circle. - **Area of the Full Circle:** \( \pi r^2 = \pi \times 7^2 = 49\pi \). - **Quarter of the Circle:** \( \frac{49\pi}{4} \). 4. **Combine Both Areas:** \[ 21 + \frac{49\pi}{4}. \] Therefore, the value of the integral is: \[ 21 + \frac{49\,\pi}{4}. \] **Final Answer:** After simplifying, the integral is equal to 21 plus 49⁄4 times π. Thus, 21 + 49 π⁄4