The Fundamental Theorem of Calculus states that the integral of a rate of change of a quantity gives the net change in that quantity. Here, $$r'(t)$$ is the rate of change of the radius. Therefore, the total increase (net change) in the radius from $$t=2$$ to $$t=6$$ is given by the definite integral $$\int_2^6 r'(t) dt$$.

The initial number of orders is 500. The total number of orders processed from $$t=0$$ to $$t=4$$ is the integral of the rate of processing, $$\int_0^4 P(t) dt$$. The number of orders remaining is the initial amount minus the total amount processed. Thus, the expression is $$500 - \int_0^4 (20t + 50) dt$$.

Let $$M(t)$$ be the amount of the drug in the bloodstream. The rate of change of this amount is $$M'(t) = A(t) - E(t)$$. To find a maximum value for $$M(t)$$, we must find a critical point by setting its derivative to zero: $$M'(t) = A(t) - E(t) = 0$$. This occurs when $$A(t) = E(t)$$, where the rate of change switches from positive to negative.

The value of $$Q(1)$$ is the initial value $$Q(0)$$ plus the net change from $$t=0$$ to $$t=1$$. $$Q(1) = Q(0) + \int_0^1 Q'(t) dt = 5 + \int_0^1 \frac{1}{t^2+1} dt = 5 + [\arctan(t)]_0^1 = 5 + (\arctan(1) - \arctan(0)) = 5 + (\frac{\pi}{4} - 0) = 5 + \frac{\pi}{4}$$.

The net rate of change of people in the park is $$R(t) = E(t) - L(t) = (100 + 20t) - 80 = 20 + 20t$$. The net change from $$t=0$$ to $$t=3$$ is the integral of the net rate: $$\int_0^3 (20 + 20t) dt = [20t + 10t^2]_0^3 = (20(3) + 10(3^2)) - 0 = 60 + 90 = 150$$ people.

The amount of fuel remaining is the initial amount minus the total amount of fuel burned. The amount burned is the integral of the rate of consumption from $$t=0$$ to $$t=100$$. Thus, the remaining fuel is $$50000 - \int_0^{100} 150\sqrt{t} dt$$.

If the amount of snow is the same at $$t=24$$ as at $$t=0$$, then the net change in snow is zero. The net change equals the integral of the net rate of change: $$\int_0^{24} (S(t) - M(t)) dt = 0$$. This doesn't require that $$S(t) = M(t)$$ at every instant, only that the total snow added equals the total snow melted over the entire period.

The question asks for the average rate of change of the number of cars. Since $$C(t)$$ is the total number of cars (an accumulation function), the rate is $$C'(t)$$. The average value of the rate $$C'(t)$$ over $$[2, 5]$$ is $$\frac{1}{5-2}\int_2^5 C'(t)dt$$. By the Fundamental Theorem of Calculus, this is equal to $$\frac{C(5)-C(2)}{5-2}$$. This is also the standard definition of average rate of change for the function $$C(t)$$.

The amount of snow that fell from $$t=2$$ to $$t=6$$ is given by $$S(6) - S(2) = \int_2^6 s(x) dx$$. $$\int_2^6 (1.5 + \sin(\frac{\pi x}{6})) dx = [1.5x - \frac{6}{\pi}\cos(\frac{\pi x}{6})]_2^6 = (1.5(6) - \frac{6}{\pi}\cos(\pi)) - (1.5(2) - \frac{6}{\pi}\cos(\frac{\pi}{3})) = (9 - \frac{6}{\pi}(-1)) - (3 - \frac{6}{\pi}(\frac{1}{2})) = (9 + \frac{6}{\pi}) - (3 - \frac{3}{\pi}) = 6 + \frac{9}{\pi}$$ cm.

The change in profit from selling the 101st unit through the 200th unit corresponds to the accumulation of profit from $$x=100$$ to $$x=200$$. The total change is the definite integral of the marginal profit function over this interval, which is $$\int_{100}^{200} (150 - 0.2x) dx$$.