적분의 활용 전체 문제

Force is mass $\times$ acceleration (with acceleration equal to $g$ in this problem), and both in this scenario are constant, so we don't need an integral–only a product–to calculate the force acting on the block.

To find the force in newtons, recall one newton is one $\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}$, so we need the mass of our block in kg. Specifically, our block has mass $\frac{3}{1000}$ kg. So, the force involved is $$F=\left(\frac{3}{1000}\text{ kg}\right)\times\left(9.8~\frac{\text{m}}{\text{sec}^2}\right) = 0.0294~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2} = 0.0294~\text{N}$$

To find the work in joules, recall one joule is one newton-metre: that is, one newton of force acting over one metre. So, we need our distance in metres. $$W = \big( 0.0294~\text{N}\big)\times\left(\frac{1}{10}~\text{m}\right) = 0.00294~\text{N}\cdot\text{m} = 0.00294~\text{J}$$

The force of the rock is one newton, or one kilogram-metre per second squared, so $$1~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2} = (x~\text{kg})\left(9.8~\frac{\text{m}}{\text{sec}^2}\right)$$ Therefore, the mass of the rock is $\frac{1}{9.8}$ kg, or about 102 grams.

Now, since one joule is one newton-metre, the amount of work required to counteract 1 N of gravitational force for one metre is precisely one joule.

Remark: having an idea of how much work a joule is, and how much force a newton is, is a good tool for checking the reasonableness of your work. For example, after this question, if you calculate that a marble weighs 100 N, you can be pretty sure there's an error in your calculation.

(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.

Hooke's law says that the force required to stretch a spring $x$ units past its natural length is proportional to $x$; that is, there is some constant $k$ associated with the individual spring such that the force required to stretch it $x$ m past its natural length is $kx$.

• Solution 1: Since the force required to stretch the spring is proportional to the amount stretched, and the force acting on the spring is proportional to the mass hanging from it, we conclude the amount the spring stretches is proportional to the mass hung from it. So, if 1 kg stretches it 1 cm, then 10 kg will stretch it 10 cm. We should mark the wall 10 cm below the bottom of the spring as it hangs unloaded.
• Solution 2: We can find $k$ from the test with the bag of water. The force exerted by the bag of water was $(1\text{ kg})(9.8\text{ m/sec}^2) = 9.8$ N$=k(1 \text{ cm})$. So, $$k = \frac{9.8~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}}{0.01\text{ m}} = 980 \frac{\text{kg}}{\text{sec}^2}$$

If we hang 10 kg from the spring, gravity exerts a force of $(10 \text{ kg})(9.8 \text{ m/sec}^2) = 98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}$. This will be matched by the spring with a force of $kx$ newtons, where $k$ is the spring constant and $x$ is the amount stretched.

$$\begin{aligned} kx&=98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\\ \left(980 \frac{\text{kg}}{\text{sec}^2}\right)\left(x\text{ m}\right)&=98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\\ x&=\frac{1}{10}\text{ m} = 10\text{ cm} \end{aligned}$$

So, we should put the mark at 10 cm below the natural length of the spring.

(a) The entire interval has length $b-a$, and we're cutting it into $n$ pieces, so the length of one piece (and hence the distance between two consecutive samples) is $\frac{b-a}{n}$. (b) The first sample, as given in the question statement, is taken at $x=a$. The second sample, then, is at $x=a+\frac{b-a}{n}$, this third is at $x=1+2\frac{b-a}{n}$, and the fourth is at $a+3\frac{b-a}{n}$. (c) The $y$-value of the fourth sample is simply $f\left(a+3\frac{b-a}{n}\right)$. Note this is the number we use in our average, not the $x$-value. (d) Our samples are $f(a)$, $f\left(a+\frac{b-a}{n}\right)$, $f\left(a+2\frac{b-a}{n}\right)$, $f\left(a+3\frac{b-a}{n}\right)$, etc. Since there are $n$ of them, we divide their sum by $n$. So, the average is:

$$\begin{aligned} &~\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+ f\left(a+2\frac{b-a}{n}\right)+\cdots + f\left(a+(n-1)\frac{b-a}{n}\right)}{n}\\ &=\frac{1}{n}\left[f(a)+f\left(a+\frac{b-a}{n}\right)f\left(a+2\frac{b-a}{n}\right)+\cdots + f\left(a+(n-1)\frac{b-a}{n}\right)\right]\\ &= \frac{1}{n}\sum_{i=1}^nf\left(a+(i-1)\frac{b-a}{n}\right) \\ \text{Remark: if we multiply and divide by $b-a$, we see this expression is equivalent to a left Riemann sum, divided by the length of our interval.} \\ &= \frac{1}{b-a}\sum_{i=1}^nf\left(a+(i-1)\frac{b-a}{n}\right)\frac{b-a}{n}\\ &= \frac{1}{b-a}\sum_{i=1}^nf\left(a+(i-1)\Delta x\right)\Delta x \end{aligned}$$

As $n$ gets larger and larger, using the definition of a definite integral, this expression gets closer and closer to $\frac{1}{b-a}\int_a^bf(x)\,\mathrm{d}x$. This is one way of justifying our definition of an average of a function on an interval.

(a) Yes, the average of $f(x)$ is less than or equal to the average of $g(x)$ on $[0,10]$. The reason is that, if $f(x) \leq g(x)$ for all $x$ in $[0,10]$, then: $$\frac{1}{10}\int_0^{10}f(x)\,\mathrm{d}x \leq \frac{1}{10}\int_0^{10}g(x)\,\mathrm{d}x.$$ (b) There is not enough information to tell. It's certainly possible: for instance, take $f(x)=0$ and $g(x)=1$ for all $x$ in $[0,10]$. Then $f(x) \leq g(x)$ and the average of $f(x)$ is 0, which is less than 1, the average of $g(x)$.

However, consider $f(x) = \begin{cases} 100 & \text{ if } 0 \le x \le 0.01\\ 0 & \text{ else } \end{cases}$ and $g(x) = 0$. Then $f(x) \le g(x)$ for all $x$ in $[0.01,10]$, but the average of $f(x)$ is $0.1$, while the average of $g(x)$ is 0.

Note $-x^2+2x+1 = 2-(x-1)^2$. So, both parabolas are symmetric about the line $x=1$, and the $x$-coordinate of the centroid is $x=1$.

The parabolas meet when:

$$\begin{aligned} (x-1)^2&=2-(x-1)^2\\ 2(x-1)^2&=2\\ |x-1|&=1\\ x=0,\quad x&=2 \end{aligned}$$

At both these points, $y=1$, so we see the figure is symmetric about the line $y=1$. Then the $y$-coordinate of the centroid is $1$.

Therefore, the centroid is at $(1,1)$.

The circle and the cut-out rectangle are symmetric about the $x$-axis, and about the $y$-axis, so the centroid is the origin.

Remark: the centroid of a region doesn't have to be a point in the region!

(a) If we were to set this figure on a pencil lined up along the vertical line $x=a$, it seems pretty clear that it would fall to the left. So, the centre of mass is to the left of the line $x=a$. The same is true in (b): the added density on the left makes it only more lopsided. However, in (c), the right side is denser than the left, which could counterbalance the left. Without knowing more about the dimensions and the density, we can't say where the centre of mass is in relation to the line $x=a$.

(d) Consider a section of the figure, consisting of all points $(x,y)$ in the figure with $b\le x\le c$, and its “mirror" section on the other side of the line $x=a$. These two sections, which are drawn in red in the sketches below, will have the same area, at the same distance from $x=a$. Since we only care about the $x$-coordinate of the centre of mass, it doesn't matter that the two halves are at different $y$-coordinates. The centre of mass falls along the line $x=a$.

(e) There is the same amount of area to the left and right of the line $x=a$, as in part (d). However, the area to the right is “stretched out" more, so that it occupies space farther away from the line $x=a$. So, the centre of mass will be to the right of the line $x=a$.

• The volume of water in Tank A is $\frac{4}{3}\pi(1)^3 = \frac{4}{3}\pi$ cubic metres.
• The mass of water is $\frac{4000}{3}\pi$ kg.
• By symmetry, the centre of mass of the water when it fills Tank $A$ is exactly in the centre of the sphere, at height $\bar y_1=4$ metres above the ground (one metre above the bottom of Tank $A$, which is three metres above the ground).
• When the water is entirely in Tank $B$, its height is $\frac{2}{3}\pi$ metres. (The base of Tank $B$ has area 2 m$^2$, and the volume of water is $\frac{4}{3}\pi$ m$^3$.) By symmetry, the centre of mass is exactly halfway up, at height $\bar y_2=\frac{1}{3}\pi$ metres.
• So, the point mass in our model is moved from $\bar y_1=4$ to $\bar y_2=\frac{1}{3}\pi$, a distance of $4-\frac{1}{3}\pi$ metres, by gravity.
• The work involved is: $$\left(\frac{4000}{3}\pi~\text{kg}\right)\times\left(4-\frac{1}{3}\pi~\text{m}\right)\times\left(9.8~\frac{\text{m}}{\text{sec}^2}\right) = \frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}$$