적분의 활용 전체 문제

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.

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 $u=a+b-x$, $\mathrm{d}x=-\mathrm{d}x$:

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

(a) The function $A(x)$ only gives us information about an integral when one limit of integration is zero. We can get around this by using properties of definite integrals from Section 1.2 to break our integral into two integrals, each of which has 0 as one limit of integration. So, we find the average of $f(t)$ on $[a,b]$ as follows:

$$\begin{aligned} \frac{1}{b-a}\int_a^b f(t)\,\mathrm{d}t&= \frac{1}{b-a}\left(\int_a^0 f(t)\,\mathrm{d}t + \int_0^b f(t)\,\mathrm{d}t\right)\\ &=\frac{1}{b-a}\left(-\int_0^a f(t)\,\mathrm{d}t + \int_0^b f(t)\,\mathrm{d}t \right)\\ &=\frac{1}{b-a}\left(-a\cdot\underbrace{\frac{1}{a}\int_0^a f(t)\,\mathrm{d}t}_{A(a)} + b\cdot\underbrace{\frac{1}{b}\int_0^b f(t)\,\mathrm{d}t}_{A(b)} \right)\\ &=\frac{1}{b-a}\left(-aA(a)+bA(b) \right) = \frac{bA(b) - aA(a)}{b-a} \end{aligned}$$ (b) From the definition of $A(x)$, we know $$A(x) = \frac{1}{x}\int_0^x f(t)\,\mathrm{d}t$$ That is, $$xA(x) =\int_0^x f(t)\,\mathrm{d}t$$ To find $f(x)$, we differentiate both sides. For the left side, we use the product rule; for the right side, we use the Fundamental Theorem of Calculus part 1. $$A(x)+xA'(x) = f(x)$$ So, $f(t)=A(t)+tA'(t)$.