적분의 활용 전체 문제

12.5 ft–lb

5/3 ft–lb

0.2625 = 21/80 J

$f\cdot d/2$ J

45 ft–lb

5 ft–lb

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.

According to the formula for gravity between two objects, the earth and moon will gravitationally attract one another no matter how far apart they are, so what we're looking for is the work to separate them infinitely far. That is, we want to calculate $\displaystyle\int_a^\infty F(r)\,\mathrm{d}r$, where $a = 400\,000\,000$ m.

If we take $m_1$ and $m_2$ to be the mass of the earth and moon as given in the question statement, then:

$$\begin{aligned}Gm_1m_2 &= \left(6.7\times 10^{-11} \ \frac{\text{m}^3}{\text{kg}\cdot\text{sec}^2}\right) \left(6 \times 10^{24}\text{ kg}\right)\left(7 \times 10^{22}\text{ kg}\right)\\ &\approx 2.8 \times 10^{37} ~\frac{\text{kg}\cdot\text{m}^3}{\text{sec}^2} \end{aligned}$$

With that out of the way, let's calculate our work.

$$\begin{aligned} W&=\int_a^\infty F(r)\,\mathrm{d}r =\lim_{b \to \infty} \int_a^b G\dfrac{m_1m_2}{r^2}\,\mathrm{d}r \\&= (Gm_1m_2)\lim_{b \to \infty}\left[-\frac{1}{r}\right]_a^b \\ &=(Gm_1m_2)\lim_{b \to \infty} \left(\frac{1}{a}-\frac{1}{b}\right)\\&= \frac{Gm_1m_2}{a}\\ &\approx\frac{2.8 \times 10^{37} ~\frac{\text{kg}\cdot\text{m}^3}{\text{sec}^2}}{4\times 10^8 \text{ m}}\\ &=7 \times 10^{28} \text{ J} \end{aligned}$$

Remark: since the force of gravity between the earth and the moon gets weaker as they are farther apart, it takes less and less work to move them each kilometre. If we move them a finite distance apart, the work involved will always be less than $7\times 10^{28}$ joules, no matter how huge that finite distance is. If we move them a very, very long (but finite) distance apart, the work we did will be quite close to (but still less than) $7\times 10^{28}$ joules.

(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.

By definition, the average is

$$\begin{aligned} \frac{1}{e-1}\int_1^e x^2\log x\ \mathrm{d}x & \\ \text{To antidifferentiate, we use integration by parts with $u=\log x$ and $\mathrm{d}v=x^2\,\mathrm{d}x$, hence $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=\frac{1}{3}x^3$.} \\ \frac{1}{e-1}\int_1^e x^2\log x\ \mathrm{d}x&= \frac{1}{e-1}\left(\left[\frac{1}{3}x^3\log x\right]_1^e - \int_1^e \frac{1}{3}x^2\,\mathrm{d}x \right) \\&= \frac{1}{e-1}\bigg[\frac{x^3}{3}\log x-\frac{x^3}{9}\bigg]_{x=1}^{x=e}\\ &=\frac{1}{e-1}\bigg[\frac{e^3}{3}-\frac{e^3}{9}+\frac{1}{9}\bigg]\\ &=\frac{1}{e-1}\bigg[\frac{2}{9}e^3+\frac{1}{9}\bigg] \end{aligned}$$

By definition, the average value in question equals

$$\frac1{\pi/2-0} \int_0^{\pi/2} ( 3\cos^3x + 2\cos^2x ) \,\mathrm{d}x = \frac2\pi \bigg( \int_0^{\pi/2} 3\cos^3x \,\mathrm{d}x + \int_0^{\pi/2} 2\cos^2x \,\mathrm{d}x \bigg)$$

For the first integral we use the substitution $u = \sin x$, $\mathrm{d}u=\cos x\,\mathrm{d}x$, $\cos^2 x=1-\sin^2 x = 1-u^2$. Note that the endpoints $x=0$ and $x=\frac\pi2$ become $u=0$ and $u=1$, respectively.

$$\begin{aligned} \int_0^{\pi/2} 3 \cos^3x \,\mathrm{d}x &= \int_0^{\pi/2} 3 \cos^2 x \cos x \,\mathrm{d}x \\ &= \int_0^{1} 3(1-u^2) \,\mathrm{d}u \\ &= (3u-u^3)\Big|_0^1 = 2. \end{aligned}$$

For the second integral we use the trigonometric identity $\cos^2x\,\mathrm{d}x = \frac{1 + \cos(2x)}{2}$.

$$\begin{aligned} 2 \int_0^{\pi/2} \cos^2x\,\mathrm{d}x &= \int_0^{\pi/2} \big(1 + \cos(2x) \big) \,\mathrm{d}x \\ &= \bigg[ x + \frac{1}{2}\sin(2x) \bigg]_0^{\pi/2} = \frac\pi2 \end{aligned}$$

Therefore, the average value in question is

$$\frac2\pi \bigg( \int_0^{\pi/2} 3\cos^3x \,\mathrm{d}x + \int_0^{\pi/2} 2\cos^2x \,\mathrm{d}x \bigg) = \frac2\pi \bigg( 2 + \frac\pi2 \bigg) = \frac{4}{\pi}+1.$$

By definition, the average value in question equals

$$\text{Ave}=\frac1{\pi/k-0} \int_0^{\pi/k} \sin(kx) \,\mathrm{d}x$$

To evaluate the integral, we use the substitution $u = kx$, $\mathrm{d}u=k\,\mathrm{d}x$. Note that the endpoints $x=0$ and $x=\pi/k$ become $u=0$ and $u=\pi$, respectively. So

$$\begin{aligned} \text{Ave}=\frac{k}{\pi} \int_0^\pi \sin(u) \,\frac{\mathrm{d}u}{k} =\frac{1}{\pi}\Big[-\cos(u)\Big]_0^\pi =\frac{2}{\pi} \end{aligned}$$

Remark: the average does not depend on $k$. To see why this is, note that $\sin(kx)$ runs between $-1$ and $1$ as $x$ changes. When $x=0$, $kx=0$, and when $x=\pi/k$, $kx=\pi$. So, our function $\sin (kx)$ runs exactly from $\sin 0 =0$ to $\sin(\pi/2)=1$, then back down to $\sin\pi=0$.

By definition, the average value is

$$\begin{aligned} \frac{1}{e-1}\int_1^e\frac{\log x}{x}\ \mathrm{d}x& \\ \text{To integrate, we use the substitution $u=\log x$, $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$. Then the limits of integration become 0 and 1, respectively.} \\ \frac{1}{e-1}\int_1^e\frac{\log x}{x}\ \mathrm{d}x& =\frac{1}{e-1}\int_0^1u\ \mathrm{d}u =\frac{1}{e-1}\bigg[\frac{u^2}{2}\bigg]_0^1 =\frac{1}{2(e-1)} \end{aligned}$$

By definition, the average value is:

$$\begin{aligned} \frac{1}{2\pi}\int_0^{2\pi}\cos^2x\,\mathrm{d}x =\frac{1}{2\pi}\cdot\frac{1}{2}\int_0^{2\pi}\big(\cos(2x)+1\big)\,\mathrm{d}x =\frac{1}{4\pi}\Big[\frac{\sin(2x)}{2}+x\Big]_0^{2\pi} =\frac{1}{4\pi}\cdot 2\pi =\frac{1}{2} \end{aligned}$$

The question tells you $\frac{1}{1-0}\int_0^1 f(x)\,\mathrm{d}x = \frac{f(0)+f(1)}{2}$. Note $f(0)=c$, and $f(1) = a+b+c$.

$$\begin{aligned} \frac{1}{1-0}\int_0^1 f(x)\,\mathrm{d}x &= \frac{f(0)+f(1)}{2} = \frac{c+(a+b+c)}{2}\\ \int_0^1 \big(ax^2+bx+c \big)\,\mathrm{d}x&=\frac{a+b+2c}{2}\\ \left[\frac{a}{3}x^3+\frac{b}{2}x^2+cx \right]_0^1&=\frac{a}{2}+\frac{b}{2}+c \\\frac{a}{3}+\frac{b}{2}+c&=\frac{a}{2}+\frac{b}{2}+c\\ \frac{a}{3}&=a\\ a&=0 \end{aligned}$$

That is, $f(x)$ is linear.

The function $g(x)=f(a+b-x)$, on the interval $[a,b]$, is a mirror of the function $f(x)$, with $g(a)=f(b)$ and $g(b)=f(a)$. So, $\int_a^b f(a+b-x)\,\mathrm{d}x = \int_a^bf(x)\,\mathrm{d}x$, and hence $\frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x = \frac{1}{b-a}\int_a^bf(x)\,\mathrm{d}x$, so the average value of $f(a+b-x)$ on $[a,b]$ is $A$.

Alternately, we can evaluate $\frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x$ directly, using the substitution $u=a+b-x$, $\mathrm{d}x=-\mathrm{d}x$:

$$\begin{aligned} \frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x&=\frac{-1}{b-a}\int_{u(a)}^{u(b)} f(u)\,\mathrm{d}u\\ &=\frac{-1}{b-a}\int_{b}^{a} f(u)\,\mathrm{d}u \\&=\frac{1}{b-a}\int_{a}^{b} f(u)\,\mathrm{d}u \\&=A \end{aligned}$$