적분법 전체 문제

The intervals of our rectangles are $[0,\frac{\pi}{4}]$, $[\frac{\pi}{4},\frac{\pi}{2}]$, $[\frac{\pi}{2},\frac{3\pi}{4}]$, and $[\frac{3\pi}{4},\pi]$. Since we're taking a left Riemann sum, we find the height of the rectangles at the left endpoints of the intervals.

• $x = 0$: The distance from $\cos 0$ to $\sin 0$ is 1, so our first rectangle has height 1.
• $x = \frac{\pi}{4}$: The distance from $\cos \frac{\pi}{4}$ to $\sin \frac{\pi}{4}$ is 0, so our second rectangle has height 0.
• $x = \frac{\pi}{2}$: The distance from $\cos \frac{\pi}{2}$ to $\sin \frac{\pi}{2}$ is 1, so our third rectangle has height 1.
• $x = \frac{3\pi}{4}$: The distance from $\cos \frac{3\pi}{4}$ to $\sin \frac{3\pi}{4}$ is $\sin(3\pi/4)-\cos(3\pi/4) =\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)=\sqrt{2}$, so our fourth rectangle has height $\sqrt{2}$.

So, our approximation for the area between the two curves is $$\frac{\pi}{4}\left(1+0+1+\sqrt{2}\right)=\frac{\pi}{4}\left(2+\sqrt{2}\right)$$

(a) We are finding the area in the interval from $x=0$ to $x=\frac{\pi}{2}$. Since we're taking $n=5$ rectangles, our rectangles cover the following intervals: $$\left[0,\frac{\pi}{10}\right],\quad \left[\frac{\pi}{10},\frac{\pi}{5}\right],\quad \left[\frac{\pi}{5},\frac{3\pi}{10}\right],\quad \left[\frac{3\pi}{10},\frac{2\pi}{5}\right],\quad \left[\frac{2\pi}{5},\frac{\pi}{2}\right].$$

(b) We are finding the area in the interval from $y=0$ to $y=\frac{\pi}{2}$. (In general, when we switch from horizontal rectangles to vertical, the limits of integration will change–it's only coincidence that they are the same in this example.) Since we're taking $n=5$ rectangles, these rectangles cover the following intervals of the $y$-axis: $$\left[0,\frac{\pi}{10}\right],\quad \left[\frac{\pi}{10},\frac{\pi}{5}\right],\quad \left[\frac{\pi}{5},\frac{3\pi}{10}\right],\quad \left[\frac{3\pi}{10},\frac{2\pi}{5}\right],\quad \left[\frac{2\pi}{5},\frac{\pi}{2}\right].$$

The question doesn't specify which endpoints we're using. Let's use upper endpoints, to match part (a).

The area between the curve $y= \frac{1}{(2x-4)^2}$ and the $x$-axis, with $x$ running from $a=0$ to $b=1$, is exactly the definite integral of $\frac{1}{(2x-4)^2}$ with limits $0$ and $1$.

$$\begin{aligned} \text{Area}&=\int_0^1 \frac{\mathrm{d}x}{(2x-4)^2}&u=2x-4,\quad\mathrm{d}u=2\ \mathrm{d}x\\ &=\frac{1}{2}\int_{-4}^{-2}\frac{1}{u^2}\mathrm{d}u = \frac{1}{2}\left[\frac{-1}{u}\right]_{u=-4}^{u=-2}\\ &=\frac{1}{2}\Big[\frac{1}{2}-\frac{1}{4}\Big] =\frac{1}{8} \end{aligned}$$

We will compute the area by using thin vertical strips, as in the sketch below:

By looking at the sketch above, we guess the line $y = 4 + 2\pi - 2x$ intersects the curve $y = 4 + \pi \sin x$ when $x=\frac{\pi}{2},$ $x=\pi$, and $x=\frac{3\pi}{2}$. Let's make sure these are correct by plugging them into the two equations, and making sure the $y$-values match:

$$\begin{array}{|c|c|c|c|}\hline x & 4+2\pi-2x & 4+\pi\sin(x) & \text{match?} \\ \hline \frac{\pi}{2} & 4+\pi & 4+\pi & \text{\checkmark} \\ \hline \pi & 4 & 4 & \text{\checkmark} \\ \hline \frac{3\pi}{2} & 4-\pi & 4-\pi & \text{\checkmark} \\ \hline \end{array}$$

Also from the sketch, we see that:

• When $\frac{\pi}{2} \le x \le \pi$, the top of the strip is at $y = 4 + \pi \sin x$ and the bottom of the strip is at $y = 4 + 2\pi - 2x$. So the strip has height $\big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]$ and width $\mathrm{d}x$, and hence area $\big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]\mathrm{d}x$.
• When $\pi \le x \le \frac{3\pi}{2}$, the top of the strip is at $y = 4 + 2\pi - 2x$ and the bottom of the strip is at $y = 4 + \pi \sin x$. So the strip has height $\big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]$ and width $\mathrm{d}x$, and hence area $\big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]\mathrm{d}x$.

Now we can calculate:

$$\begin{aligned} \text{Area} &= \int_{\pi/2}^\pi \big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]\ \mathrm{d}x +\int^{3\pi/2}_\pi \big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]\ \mathrm{d}x\\ &= \int_{\pi/2}^\pi \big[\pi \sin x- 2\pi + 2x\big]\ \mathrm{d}x +\int^{3\pi/2}_\pi \big[2\pi - 2x- \pi \sin x\big]\ \mathrm{d}x\\ &=\Big[-\pi \cos x- 2\pi x + x^2\Big]_{\pi/2}^\pi +\Big[2\pi x - x^2+ \pi \cos x\Big]^{3\pi/2}_\pi\\ &=\left[\pi-\pi^2+\frac{3}{4}\pi^2\right] +\left[\pi^2-\frac{5}{4}\pi^2+\pi\right]\\ &=2\Big[\pi-\frac{1}{4}\pi^2\Big] \end{aligned}$$

The columns have the same volume. We can see this by chopping up the columns into horizontal cross-sections. Each cross-section has the same area as the cookie cutter, $A$, and height $\mathrm{d}y$. Then in both cases, the volume of the column is $$\int_{0}^h A~\mathrm{d}y = hA \text{ cubic units}$$

Notice $f(x)$ is a piecewise linear function, so we can find explicit equations for each of its pieces from the graph. The radii will be determined by the $x$-values, so below we give the $x$-values as functions of $y$.

If we imagine rotating the region from the picture about the $y$-axis, there will be two kinds of washers formed: when $y<1$, we have a “double washer," two concentric rings. When $y>1$, we have a single ring.

• Washers when $\mathbf{1<y \le 6}$: If $y>1$, then our washer has inner radius $2+\frac{2}{3}y$, outer radius $6-\frac{2}{3}y$, and height $\mathrm{d}y$.

• Washers when $\mathbf{0\le y <1}$: When $0 \le y < 1$, we have a “double washer," two concentric rings corresponding to the two “humps" in the function. The inner washer has inner radius $r_1=y$ and outer radius $R_1=2-y$. The outer washer has inner radius $r_2=2+\frac{2}{3}y$ and outer radius $R_2=6-\frac{2}{3}y$. The thickness of the washers is $\mathrm{d}y$.

(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.

(a) In the following plot, we see that the model is reasonable because it appears to be periodic and repeats every $24$ hours, as well as that the rate of consumption seems to peak at the times of day where people are most active in their homes: around $7$ a.m. and around $5$ p.m. (b) The meaning of $\int_0^{24} r(t) \, dt$ is the total power consumed in $24$ hours, measured in megawatt-hours. This is because $r(t)$ is measured in megawatts and $\triangle t$ in hours, and thus in the Riemann sum that corresponds to the definite integral, the terms in the sum of the form $r(t) \cdot \triangle t$ have units “megawatt-hours”. Moreover, by integrating the rate of power consumption we get the total change in power consumed. (c) In a typical day, we know that Mathland consumes $\int_0^{24} r(t) \, dt$ megawatt-hours of power, and thus we seek to evaluate the integral $$\int_0^{24} r(t) \, dt = \int_0^{24} \left[ 4 + \sin(0.263t + 4.7) + \cos(0.526t+9.4) \right] \, dt\,.$$ To do so, we use the subtitutions $u = 0.263t + 4.7$ (so $du = 0.263 \, dt$) and $w = 0.526t + 9.4$ (so $dw = 0.526 \, dt$) to help us antidifferentiate the two trigonometric expressions, and thus we find that $$\int \sin(0.263t + 4.7) \, dt = \frac{1}{0.263} \int \sin(u) \, du = -\frac{1}{0.263} \cos(u) = -\frac{1}{0.263} \cos(0.263t + 4.7)$$ and $$\int \cos(0.526t + 9.4) \, dt = \frac{1}{0.526} \int \cos(w) \, dw = \frac{1}{0.526} \sin(w) = \frac{1}{0.526} \sin(0.526t + 9.4)\,.$$ Putting this all together and using the Fundamental Theorem of Calculus, we have $$\begin{aligned}\int_0^{24} r(t) \, dt &= \left. 4t -\frac{1}{0.263} \cos(0.263t + 4.7) + \frac{1}{0.526} \sin(0.526t + 9.4) \right|_0^{24} \\ &= \left[ 4 \cdot 24 -\frac{1}{0.263} \cos(0.263 \cdot 24 + 4.7) + \frac{1}{0.526} \sin(0.526 \cdot 24 + 9.4) \right] - \left[ -\frac{1}{0.263} \cos(4.7) + \frac{1}{0.526} \sin(9.4) \right] \\ &\approx 95.7809\end{aligned}$$ megawatt-hours of power used in $24$ hours. (d) We know that the average value of the rate function $r$ on $[0,24]$ is $$\displaystyle r_{\operatorname{AVG} [0,24]} = \frac{1}{24-0} \int_{0}^{24} r(t) \, dt\,.$$ Thus, by our work in (c), we have that $$\displaystyle r_{\operatorname{AVG} [0,24]} \approx \frac{1}{24} \cdot 95.7809 = 3.99087$$ which is measured in megawatts.