Final answer:
The integral of 25t^2 evaluated from 5 to 0 is -1041.67. It represents the negative area under the curve of the function 25t^2 from t=5 to t=0.
Step-by-step explanation:
The integral \(\int_{5}^{0} 25t^2 dt\) requires finding the antiderivative of 25t^2 and evaluating it from t=5 to t=0. The antiderivative of 25t^2 is \(\frac{25}{3}t^3\). Plugging in the limits, we evaluate \(\frac{25}{3}(0)^3 - \frac{25}{3}(5)^3\), which gives us -\frac{25}{3}(125) or -1041.67. Notice that the integral denotes the area under the curve, and in this context, it's important to remember that the area calculated will be negative as we are integrating from a greater to a lesser bound.