Evaluate the integral by interpreting it in terms of areas. \[ \int_{-5}^{5}\left(2 x-\sqrt{25-x^{2}}\right) d x \]

To evaluate the integral \[ \int_{-5}^{5}\left(2 x-\sqrt{25-x^{2}}\right) d x, \] we can interpret the two components of the integrand separately in terms of areas. 1. **Area Under \(2x\)**: The function \(y = 2x\) describes a straight line through the origin with a slope of 2. The area under this line from \(-5\) to \(5\) is a trapezoid since the line crosses the x-axis. - At \(x = -5\), \(y = -10\) and at \(x = 5\), \(y = 10\). Therefore, the area can be calculated as: \[ A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times (10 - (-10)) = 0. \] The area above the x-axis cancels with the area below since it’s symmetric about the origin. 2. **Area Under \(\sqrt{25 - x^2}\)**: The function \(y = \sqrt{25 - x^2}\) describes the upper half of a circle with radius 5 centered at the origin. The area of a full circle is given by \(\pi r^2\). Thus, for a radius of 5: \[ A_2 = \frac{1}{2} \times \pi \times 5^2 = \frac{25\pi}{2}. \] Now we can combine our findings. The integral represents the area under \(2x\) (which is 0) minus the area under the semicircle (\(\frac{25\pi}{2}\)): \[ \int_{-5}^{5}\left(2 x-\sqrt{25-x^{2}}\right) d x = 0 - \frac{25\pi}{2} = -\frac{25\pi}{2}. \] Thus, the value of the integral is: \[ \boxed{-\frac{25\pi}{2}}. \]