\( A = \int _ { 0 } ^ { 2 } x ^ { 2 } - 5 d x \)

To evaluate the integral \( A = \int _ { 0 } ^ { 2 } (x ^ { 2 } - 5) \, dx \), we first break it down. The integral of \( x^2 \) from 0 to 2 gives us \( \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8}{3} \), and the integral of a constant, \( -5 \), over that interval gives \( -5x \bigg|_{0}^{2} = -10 \). Combining these, we have \( A = \frac{8}{3} - 10 = \frac{8}{3} - \frac{30}{3} = -\frac{22}{3} \). This integral helps us visualize the area under the curve of the function \( x^2 - 5 \) between 0 and 2. Since the function dips below the x-axis, the negative result tells us there's more area below than above, a clear illustration of how integral calculus not only measures areas but also helps in understanding concepts like net area and the behavior of functions over intervals. Ever curious about how integrals are applied? Think about physics! For instance, the integral of a velocity function gives you the total displacement over a given time interval. Similarly, in economics, integrals can help calculate consumer and producer surplus, showing how mathematical concepts have real-life implications, helping in practical decision-making!