정적분과 급수(구분구적법)

(a) Since we are using $4$ subintervals on $[2,5]$, it follows that $\triangle x = \frac{5-2}{4} = 0.75$. Thus, $x_0 = 2$, $x_1 = 2.75$, $x_2 = 3.5$, $x_3 = 4.25$, and $x_4 = 5$. The midpoints of the respective subintervals are thus $\overline{x_1} = \frac{2+2.75}{2} = 2.375$, $\overline{x_2} = 3.125$, $\overline{x_3} = 3.875$, and $\overline{x_4} = 4.625$. Thus, $$\begin{aligned}M_4 &= f(\overline{x_1}) \cdot \triangle x + f(\overline{x_2}) \cdot \triangle x + f(\overline{x_3}) \cdot \triangle x + f(\overline{x_4}) \cdot \triangle x \\ &= \left(f(2.375) + f(3.125) + f(3.875) + f(4.625) \right) \cdot (0.75) \\ &= \left(11.125 + 13.375 + 15.625 + 17.875 \right) \cdot (0.75) \\ &= 58 \cdot 0.75 \\ &= 43.5\end{aligned}$$ An overall sketch of the key parts of $M_4$ is seen in the figure below. (b) Observe that $f(2) = 10$ and $f(5) = 19$. The region bounded by the function, the $x$-axis, and the endpoints is a trapezoid. Recalling that the area of a trapezoid is the average of the bases times the height, we see that the bounded region has area $$A = \frac{1}{2}(10 + 19)(5-2) = \frac{87}{2}\,.$$ (c) The reason that $M_4$ gives the exact area between the line and the $x$-axis is the rectangles with heights that come from the midpoint have the same area as the trapezoids that are formed by the function values at the two endpoints of each subinterval. We can also see this by observing that the “extra area” bounded by the midpoint rectangle exactly matches the “missing area” that is absent from the midpoint rectangle, as seen in the following figure. Since these two quantities balance out, the areas of the midpoint rectangles give us the exact area under the curve. Indeed, the value of $M_n$ will give the exact area under $g$ on $[2,5]$ for any value of $n$. But neither $L_n$ nor $R_n$ will be exact: $L_n$ will underestimate the true area since every rectangle will lie completely under the function $g$, whereas $R_n$ will overestimate because every rectangle lies above the function. (d) For any linear function $g$ of the form $g(x) = mx + b$ such that $g(x) \ge 0$ on the interval of interest, the midpoint rule will give the exact value of the area bounded by the function and the $x$-axis. This is true because the area of the trapezoid formed by the two endpoints on a given subinterval will be the same as the area of the rectangle whose height is given by the function at the midpoint of the subinterval.

(a) A plot of the given data is shown in the figure below. (b) For $M_3$ with the given data, the midpoints are $\overline{x_1} = 0.3$, $\overline{x_2} = 0.9$, and $\overline{x_3} = 1.5$, and $\triangle t = 0.6$, so $$M_3 = f(0.3) \cdot 0.6 + f(0.9) \cdot 0.6 + f(1.5) \cdot 0.6 = 166 \cdot 0.6 = 99.6\,.$$ Therefore, the car travels approximately $99.6$ feet over the $1.8$ second interval it brakes. (c) Using the standard formulas, $$L_6 = f(0) \cdot 0.3 + f(0.3) \cdot 0.3 + \cdots + f(1.5) \cdot 0.3 = 380 \cdot 0.3 = 114$$ and $$R_6 = f(0.3) \cdot 0.3 + f(0.6) \cdot 0.3 + \cdots + f(1.8) \cdot 0.3 = 280 \cdot 0.3 = 84\,.$$ Thus, $$\frac{1}{2}(L_6 + R_6) = \frac{114+84}{2} = 99\,.$$ Each of these three quantities is an estimate of the total distance traveled: $L_6$ is an overestimate and $R_6$ is an underestimate. (d) The maximum possible distance the car traveled before it stopped is $114$ feet. This is because $L_n$ is always an overestimate when the function is decreasinging on the interval.