정적분의 뜻과 성질

(a) We observe that $v(t) = 0$ at $t = 0$, $t = 1$, and $t = 3$, and $v(t)$ is positive on the intervals $0 < t < 1$ and $3 < t < 4$, while $v(t)$ is negative on $1 < t < 3$. The total change in position, $P$, of the object is thus given by $$P = \int_0^1 v(t) \, dt + \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt = \int_0^4 v(t) \, dt\,.$$ (b) Using computational technology, we use the midpoint rule to estimate the total change of position, $\int_0^4 v(t) \, dt$. We find that $M_{25} = 2.6496$, $M_{50} = 2.6624$, $M_{75} = 2.66477$, and $M_{100} = 2.6656$. Since it appears that neither of the first two decimal places are changing as we increase the number of subintervals, it looks like $P = \int_0^4 v(t) \, dt \approx 2.665$ is accurate to two decimal places. (c) We noted in (a) where $v(t)$ is positive and negative. In particular, on $1 < t < 3$, $v(t) < 0$, and thus $\int_1^3 v(t) \, dt$ will give us a negative change in position; the distance traveled on this interval is therefore $-\int_1^3 v(t) \, dt$, and hence the total distance traveled, $D$, on $[0,4]$ is $$D = \int_0^1 v(t) \, dt - \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt\,.$$ (d) To use Riemann sums to estimate $D$, we have to estimate each of the three integrals individually. We find that

• $\int_0^1 v(t) \, dt \approx M_{100} = 0.41669$,
• $\int_1^3 v(t) \, dt \approx M_{100} = -2.6668$, and
• $\int_1^4 v(t) \, dt \approx M_{100} = 4.91667$ From experimenting with different $n$-values in $M_n$, it is apparent that each of these individual estimates is accurate to at least $3$ decimal places, and thus we expect their sum to be accurate to at least $2$. Hence, $$D = \int_0^1 v(t) \, dt - \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt \approx 0.41669 - (-2.6668) + 4.91667 = 8.00016\,.$$ Therefore $D \approx 8.00016$ is an estimate of distance traveled accurate to at least $2$ decimal places. (e) We know that the average velocity on $[0,4]$ is given by the integral expression $$v_{\operatorname{AVG} [0,4]} = \frac{1}{4-0} \int_0^4 v(t) \, dt\,.$$ We estimated the definite integral in (b), so using that result, $$v_{\operatorname{AVG} [0,4]} \approx \frac{1}{4-0} 2.665 = 0.66625$$ feet per second is the object's average velocity, accurate to at least two decimal places. It makes sense that the average velocity is positive since the total change in position is positive.