적분의 활용 전체 문제

F

F

T

In SI units, it is one joule, i.e., one Newton–meter, or kg$\cdot$m/s$^2\cdot$m. In Imperial Units, it is ft–lb.

The same.

Smaller.

force; distance

Answers will vary.

Answers will vary.

This seems like it might be true: if $f$ is getting closer and closer to zero, as $x$ grows towards infinity, then over time the later values will become a larger and larger portion of the total interval we're looking at, and so the average should look more and more like $f(x)$ when $x$ is large–that is, like 0. That's some intuition to start us out, but it isn't a rigorous argument. To be sure we haven't overlooked something, let's use the definition of an average to express $A(x)$.

$$\begin{aligned} A(x)&=\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t\\ \lim_{x \to \infty}A(x)&=\lim_{x \to \infty}\frac{1}{x}\int_0^x f(t)\,\mathrm{d}t \\ \text{If $\int_0^\infty f(t) \,\mathrm{d}t$ converges, then this limit is 0, and the statement is true. So, suppose it does not converge. Since $f(x)$ is positive, that means $\lim\limits_{x \to \infty} \int_0^x f(t)\,\mathrm{d}(t) = \infty$, so we can use l'H\^{o}pital's rule. To differentiate the numerator, we use the Fundamental Theorem of Calculus part 1.} \\ \lim_{x \to \infty}A(x)&=\lim_{x \to \infty}\frac{\int_0^x f(t)\,\mathrm{d}t}{x} = \lim_{x \to \infty}\frac{f(t)}{1}=0 \end{aligned}$$

So, the statement is true whether $\int_0^xf(t)\,\mathrm{d}t$ converges or not.

(a) To find the mass of the first bar, we use the fact that $M = \int_0^a \rho(x) \, dx$, with $a = 10$ and $\rho(x) = \frac{1}{1+x^2}$. Thus, the mass of the first bar is $$M_1 = \int_0^{10} \frac{1}{1+x^2} \, dx = \left. \arctan(x) \right|_0^{10}\,,$$ so $M_1 = \arctan(10) \approx 1.47113$. Similar computations for the second bar shows that its mass is $$M_2 = \int_0^{10} e^{-0.1x} \, dx = \left. -10e^{-0.1x} \right|_0^{10}\,,$$ so $M_2 = 10 - 10e^{-1} \approx 6.32121$. (b) To find the center of mass of each bar, we use the standard formula: $$\overline{x} = \frac{\int_0^{10} x \rho(x) \, dx}{\int_0^{10} \rho(x) \, dx}\,.$$ Having already evaluated the denominator for each mass distribution, we simply need to compute the numerator. It is straightforward to show that $$\int_0^{10} x \frac{1}{1+x^2} \, dx = \frac{1}{2} \ln(101) \approx 2.30756\,,$$ so the first bar's center of mass is $$\overline{x_1} = \frac{\arctan(10)}{\frac{1}{2}\ln(101)} \approx \frac{2.30756}{1.47113} \approx 1.56857\,.$$ For the second bar, $$\int_0^{10} x e^{-0.1x} \, dx = 100-200e^{-1} \approx 26.42411\,,$$ so its center of mass is $$\overline{x_2} = \frac{10 - 10e^{-1}}{100-200e^{-1}} \approx \frac{26.42411}{6.32121} \approx 4.18023\,.$$ (c) Consider a new 10 cm bar whose mass density function is $f(x) = \rho(x) + p(x)$. (i) The mass of this new bar is $M = \int_0^{10} (\rho(x) + p(x)) \, dx$. Using the additive property of the definite integral, $$M = \int_0^{10} \rho(x) \, dx + \int_0^{10} p(x) \, dx \approx 1.47113 + 6.32121 = 7.79234\,.$$ (ii) Applying the same idea to $\int_0^{10} xf(x) \, dx$, we have $$\int_0^{10} x(\rho(x) + p(x))) \, dx = \int_0^{10} x \rho(x) \, dx + \int_0^{10} x p(x) \, dx \approx 2.30756 + 26.42411 = 28.73167\,.$$ (iii) False. The center of mass of the new bar is $$\overline{x} = \frac{\int_0^{10} x(\rho(x) + p(x)) \, dx}{\int_0^{10} (\rho(x) + p(x)) \, dx} \approx \frac{2.30756 + 26.42411}{1.47113 + 6.32121} = 3.68717$$ while the average of the two bars' centers of mass is $$\frac{\overline{x_1} + \overline{x_2}}{2} \approx \frac{\frac{2.30756}{1.47113} + \frac{26.42411}{6.32121}}{2} = 2.8744\,.$$ Although the computations prove the conclusion, the result is also intuitive: the average treats the two bars as equal contributors and doesn't take into account that one bar has much more mass than the other. Indeed, a weighted average is needed to correctly compute the center of mass, which is what our earlier computations in (i) and (ii) accomplish.