적분의 활용 전체 문제

(a) $16\pi/3$ (b) $8\pi/3$ (c) $8\pi$

(a) $2\pi(\sqrt{2}-1)$ (b) $2\pi(1-\sqrt{2}+\sinh^{-1} (1))$

(a) $16\pi/3$ (b) $8\pi/3$ (c) $8\pi$ (d) $8\pi$

11,100 ft–lb

$953,284$ J

(a) 52,929.6 ft–lb (b) 18,525.3 ft–lb (c) When 3.83 ft of water have been pumped from the tank, leaving about 2.17 ft in the tank.

192,767 ft–lb. Note that the tank is oriented horizontally. Let the origin be the center of one of the circular ends of the tank. Since the radius is 3.75 ft, the fluid is being pumped to $y=4.75$; thus the distance the gas travels is $h(y)=4.75-y$. A differential element of water is a rectangle, with length 20 and width $2\sqrt{3.75^2-y^2}$. Thus the force required to move that slab of gas is $F(y) = 40\cdot45.93\cdot\sqrt{3.75^2-y^2}dy$. Total work is $\int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75-y)\sqrt{3.75^2-y^2}\ dy$. This can be evaluated without actual integration; split the integral into $\int_{-3.75}^{3.75} 40\cdot45.93\cdot(4.75)\sqrt{3.75^2-y^2}\ dy + \int_{-3.75}^{3.75} 40\cdot45.93\cdot(-y)\sqrt{3.75^2-y^2}\ dy$. The first integral can be evaluated as measuring half the area of a circle; the latter integral can be shown to be 0 without much difficulty. (Use substitution and realize the bounds are both 0.)

5.1 ft

4.1 ft

(a) We defined $\Delta x = \frac{b-a}{n}$: that is, the length of one interval, when we chop $[a,b]$ into $n$ of them. If $b$ and $a$ are measured in metres, then $\Delta x$ is measured in metres as well. So, the units of $\Delta x$ are metres.

Put another way, since $a$ and $b$ both describe a quantity in metres, $b-a$ describes a quantity in metres as well. (When we add or subtract quantities of the same units, their sum or difference is given in the same units.) Since $n$ is a unitless quantity (simply a number: not “$n$ kg" or “$n$ m"), $\frac{b-a}{n}$ still describes a quantity in metres. (If I have 6 metres of cloth, and I cut it into 3 pieces, each piece has $\frac{6}{3}=2$ metres–not 2 kilograms, or 2 metres per second.) (b) Since $F(x)$ is measured in kilogram-metres per second squared (newtons), the units of $F(x_i)$ are kilogram-metres per second squared (newtons). (c) $W$ is calculated by adding up summands of the form $F(x_i)\Delta x$. The units of $F(x_i)\Delta x$ are the products of the units of $F(x_i)$ with the units of $\Delta x$. That is, the units of $F(x_i)\Delta x$ are $\left(\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\right)\left(\text{m}\right) = \frac{\text{kg}\cdot\text{m}^2}{\text{sec}^2} = J$. The sum of terms given in joules is itself given in joules, so the units of $W$ are joules.

Let's start by converting from time spent pulling to amount pulled. When $y$ metres of rope have been pulled up, $2y$ seconds have passed, so $\frac{1}{5}y$ litres of water have leaked out of the bucket, leaving $5-\frac{1}{5}y$ litres. (This only makes sense when $\frac{1}{5}y \leq 5$, but we only consider values of $y$ from 0 to 5, so it's not a problem. That is, we're never hauling up an empty bucket that can't leak any more.)

When we've pulled up $y$ metres of rope, the mass in the bucket is $(5-\frac{1}{5}y)$ kg, so the force of gravity acting on it is $9.8(5-\frac{1}{5}y)$ N. Since we pull up 5 metres of rope, the work done is: $$W = \int_0^5 9.8\left(5-\frac{1}{5}y\right) \,\mathrm{d}y = 9.8\left[5y-\frac{1}{10}y^2\right]_0^5 = 9.8\left[25-2.5\right]=220.5\text{ J}$$

By definition, the average temperature is $$\frac{1}{3}\int_0^3 T(x)\,\mathrm{d}x =\frac{1}{3}\int_0^3 \frac{80}{16-x^2}\,\mathrm{d}x$$ We don't see an obvious substitution, but integrand is a rational function. The degree of the numerator is strictly less than the degree of the denominator, so we factor the denominator and use a partial fraction decomposition.

$$\begin{aligned} \frac{80}{16-x^2}&=\frac{80}{(4-x)(4+x)} = \frac{A}{4-x}+\frac{B}{4+x}\\ 80&=A(4+x)+B(4-x) \\ \text{ Setting $x=4$, we see $80=8A$, so $A=10$. Setting $x=-4$, we see $80=8B$, so $B=10$. } \\ \frac{1}{3}\int_0^3 \frac{80}{16-x^2}\,\mathrm{d}x&=\frac{1}{3}\int_0^3 \frac{80}{(4-x)(4+x)}\,\mathrm{d}x=\frac{1}{3}\int_0^3 \Big[\frac{ 10}{4-x}+\frac{ 10}{4+x}\Big]\,\mathrm{d}x\\ &=\frac{1}{3}\int_0^3 \Big[-\frac{10}{x-4}+\frac{10}{4+x}\Big]\,\mathrm{d}x\\ &=\frac{10}{3}\Big[-\log|x-4|+\log|x+4|\Big]_0^3 \\ &=\frac{10}{3}\log\Big|\frac{x+4}{x-4}\Big|\bigg|_0^3 =\frac{10}{3}[\log 7-\log 1] \\ &=\frac{10}{3}\log 7\quad\text{ degrees Celsius} \end{aligned}$$

Before we start answering questions, let's look at our function a little more carefully. The term $50\cos\left(\frac{t}{12}\pi\right)$ has a period of 24 hours, while the term $200\cos\left(\frac{t}{4380}\pi\right)$ has a period of one year. So, the former term describes a standard daily variation, while the latter gives a seasonal variation over the year.

(a) Using the definition of an average, the average concentration over one year ($t=0$ to $8760$) is:

$$\begin{aligned} &\frac{1}{8760}\int_0^{8760}\left(400+ 50\cos\left(\frac{t}{12}\pi\right)+ 200\cos\left(\frac{t}{4380}\pi\right)\right)\,\mathrm{d}t\\ &=\frac{1}{8760}\int_0^{8760}400\,\mathrm{d}t + \frac{50}{8760}\int_0^{8760}\cos\left(\frac{t}{12}\pi\right)\,\mathrm{d}t + \frac{200}{8760}\int_0^{8760}\cos\left(\frac{t}{4380}\pi\right)\,\mathrm{d}t\\ &= 400 + \frac{5}{876}\left[\frac{12}{\pi}\sin\left(\frac{t}{12}\pi\right)\right]_0^{8760}+ \frac{5}{219}\left[\frac{4380}{\pi}\sin\left(\frac{t}{4380}\pi\right)\right]_0^{8760} \\ \text{Since $\frac{8760}{12} = 730$, which is even, $\sin\left(\frac{8760}{12}\pi\right)=\sin(0)=0$. Also, $\sin\left(\frac{8760}{4380}\pi\right)=\sin(2\pi)=0$. } \\ &=400+ \frac{5}{876}(0) + \frac{5}{219}(0)\\ &=400\quad\text{ ppm} \end{aligned}$$

Remark: for the portions of the integral in red and blue, we also could have noticed that the integrand goes through a whole (integer) number of periods. For every period, the net signed area between the curve and the $x$-axis is zero, so we could have seen from the very beginning these terms would contribute 0 to the final average.

(b) Using the definition of an average, the average concentration over the first day ($t=0$ to $t=24$) is:

$$\begin{aligned} &\frac{1}{24}\int_0^{24} \left(400+ 50\cos\left(\frac{t}{12}\pi\right)+ 200\cos\left(\frac{t}{4380}\pi\right)\right)\,\mathrm{d}t\\ &=\frac{1}{24}\int_0^{24}400\,\mathrm{d}t + \frac{50}{24}\int_0^{24}\cos\left(\frac{t}{12}\pi\right)\,\mathrm{d}t+ \frac{200}{24}\int_0^{24}\cos\left(\frac{t}{4380}\pi\right)\,\mathrm{d}t \\ \text{Note $t=0$ to $t=24$ is one complete period for the integrand in red, so the red integral will evaluate to zero. However, $t=0$ to $t=24$ is \textit{less} than one cycle for the integrand in blue, so we expect this will contribute some non-zero quantity to the average.} \\ &=400+ 0 + \frac{200}{24}\left[\frac{4380}{\pi}\sin\left(\frac{t}{4380}\pi\right)\right]_0^{24}\\ &=400+\frac{25}{3}\cdot\frac{4380}{\pi}\sin\left(\frac{24}{4380}\pi\right)\\ &=400+\frac{25}{3}\cdot\frac{4380}{\pi}\sin\left(\frac{2}{365}\pi\right)\\ &\approx 400+199.99\\ &=599.99 \quad\text{ ppm} \end{aligned}$$

Remark: $C(0) = 400+ 50+ 200$. The red term comes from the daily variation, and over the first day this will have an average of 0. The blue term comes from the seasonal variation, which changes dramatically over the course of an entire year but won't change very much over the course of a single day. So, it is reasonable that the average concentration over the first day should be close to (but not exactly) $400+200$ ppm.

(c) The average of $N(t)$ over $[0,8760]$ is:

$$\begin{aligned} \frac{1}{8760}\int_0^{8760} \left ( 350+200\cos\left(\frac{t}{4380}\pi\right)\right)\,\mathrm{d}t&=350 + \frac{200}{8760}\bigg[\frac{4380}{\pi}\sin\left(\frac{t}{4380}\pi\right)\bigg]_0^{8760} \\&=350+\frac{200}{8760}\left[\frac{4380}{\pi}\sin\left(\frac{8760}{4380}\pi\right)\right]\\ &=350+\frac{100}{\pi}\sin\left(2\pi\right)\\&=350 \end{aligned}$$

Since the average of $C(t)$ was 400, this gives us an absolute error of $|400-350|=50$ ppm, for a relative error of $$\frac{50}{400} = 0.125,$$ or 12.5%.

That is: sampling at the same time every day, rather than throughout the day, lead to an error of 12.5% in the yearly average concentration of carbon dioxide.

(a) Here is a sketch of the specified region, which we shall call $R$.

The top of $R$ has equation $y=T(x)=e^x$, the bottom has equation $y=B(x)=-1$ and $x$ runs from $0$ to $1$. So, using vertical strips, we see that $R$ has area

$$\begin{aligned} A & = \int_0^1 \big[T(x)-B(x)\big]\,\mathrm{d}x = \int_0^1 \big[e^x-(-1)\big]\,\mathrm{d}x = \int_0^1 \big[e^x+1\big]\,\mathrm{d}x = \big[e^x+x\big]_0^1 = e \end{aligned}$$

and

$$\begin{aligned} \bar y &=\frac{1}{2A} \int_0^1 \big[T(x)^2-B(x)^2\big]\,\mathrm{d}x \\ &= \frac{1}{2e}\int_0^1 \big[ e^{2x} - 1\ \big] \,\mathrm{d}x = \frac{1}{2e} \bigg[\frac{e^{2x}}{2} - x \bigg]_0^1 \\ &=\frac{1}{2e} \bigg(\frac{e^2}{2} - 1 - \frac{1}{2} \bigg) = \frac{e}{4} - \frac{3}{4e} \end{aligned}$$

(b) To compute the volume when $R$ is rotated about the line $y=-1$

• we use thin vertical strips of width $\mathrm{d}x$ as in the figure above.

• When we rotate about the line $y=-1$, each strip sweeps out a thin disk

• whose radius is $r=T(x)-B(x)=e^x+1$ and

• whose thickness is $\mathrm{d}x$ and hence

• whose volume is $\pi r^2\,\mathrm{d}x = \pi(e^x+1)^2\,\mathrm{d}x$.

• As our leftmost strip is at $x=0$ and our rightmost strip is at $x=1$, the total volume is

$$\begin{aligned} \pi \int_0^1 (e^x+1)^2\,\mathrm{d}x &=\pi \int_0^1 (e^{2x}+2e^x+1)\,\mathrm{d}x =\pi\bigg[\frac{e^{2x}}{2}+2e^x + x\bigg]_0^1 \\ &=\pi\bigg[\Big(\frac{e^{2}}{2}+2e + 1\Big) -\Big(\frac{1}{2}+2 + 0\Big)\bigg] \\ &=\pi\Big(\frac{e^{2}}{2}+2e -\frac{3}{2}\Big) \end{aligned}$$

(a) The plot shown above (with the curve reflected across the $x$-axis to generate the solid of revolution) shows why this solid of revolution is a reasonable approximation to a baseball bat (if perhaps a bit thin between $x = 3$ and $x = 9$. (b) We use the disk method to find the volume of the solid of revolution. In the usual way, the volume of a typical slice is $$V_{\text{slice}} \approx \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \triangle x$$ and thus using a definite integral to sum the volumes of the slices, we get $$V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx\,.$$ Using technology to evaluate the integral, we find that $V \approx 52.0666$ cubic inches. (c) Given that the bat has (constant) weight density $0.6$ ounces per cubic inch, it follows that the total weight of the bat is $W \approx 0.6 \cdot 52.0666 = 31.23996$ ounces. (d) If we slice the baseball bat perpendicular to the horizontal axes, we get thin cylindrical disks as cross-sections. At a given $x$-location, the amount of weight concentrated there is approximately the weight density ($0.6$ ounces per cubic inch) times the volume of the slice, which is $V_{text{slice}} \approx \pi f(x)^2$. Thus, we see that the amount of weight concentrated at a location $x$ along the bat is given by $\rho(x) = 0.6 \pi f(x)^2$, which we can therefore view as being the weight density function for how the weight of the baseball bat is distributed from $x = 0$ to $x = 30$. (e) Even though we are dealing with “weight” instead of “mass”, we can compute the center of mass in the usual way since weight differs from mass only by the gravitational constant, and this constant is present in both the numerator and denominator of the center of mass. Thus, using the standard formula, $$\overline{x} = \frac{\int_0^{30} x \rho(x) \ dx}{\int_0^{30} \rho(x) \ dx}$$ so $$\begin{aligned}\overline{x} &= \frac{\int_0^{30} x \cdot 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx} \\ &= \frac{\int_0^{30} x \cdot (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx} \\ &\approx \frac{384.735}{16.5733} = 23.21415\end{aligned}\,,$$ so the center of mass is just over 2/3 along the length of the bat.