정적분으로 나타낸 함수

$2x\frac{|x|}{1+{x}^{2}}$

(a) Since the tide removes sand at a rate of $R(t) = 2 + 5\sin \left( \frac{4\pi t}{25} \right)$ cubic yards of sand per hour, on the time interval $[0,6]$, it follows that since the units on $R(t) \cdot \triangle t$ are “cubic yards”, the total sand removed on this time interval is $$\int_0^6 \left[2 + 5\sin \left( \frac{4\pi t}{25} \right) \right] \, dt\,.$$ (b) Because sand is being both removed and added, we can determine the total amount of sand that has been added or subtracted up to a given time $t$ by integrating the difference between the the rates at which sand is being added and removed. In particular, since the total change in sand on the beach is the integral of the rate of change, we have that the total amount of sand on the beach at time $x$ is given by the original amount (2500 cubic yards) plus the total change: $$Y(x) = 2500 + \int_0^x \left[ S(t) - R(t) \right] \, dt = 2500 + \int_0^x \left[ \frac{15t}{1+3t} - \left( 2 + 5\sin \left( \frac{4\pi t}{25} \right) \right) \right] \, dt\,.$$ (c) We know that $Y(x)$ measures the total amount of sand on the beach at time $x$. Thus, the instantaneous rate at which the total number of cubic yards of sand on the beach is changing at time $t=4$ is given by $Y'(4)$. We first compute $Y'(x)$. By the Second Fundamental Theorem of Calculus, we have $$Y'(x) = S(x) - R(x)\,,$$ and thus $Y'(4) = S(4) - R(4) \approx -1.90875$ cubic yards per hour. (d) To find where the total sand function $Y(x)$ has its minimum value on $[0,6]$, we look at the behavior of $Y'(x) = S(x) - R(x)$. Plotting $Y'(x)$ on the interval, we see that $Y'(x) = 0$ at $x \approx 5.118$, and for $0 \le x < 5.118$, $Y'(x) < 0$, and for $5.118 < x \le 6$, $Y'(x) > 0$. Thus, $Y$ changes from decreasing to increasing at $x \approx 5.118$ and has an absolute minimum there on the interval $[0,6]$. Furthermore, $Y(5.118) \approx 2492.368$.