이상적분

To find the volume of the solid, we cut it into horizontal slices, which are thin circular disks. At height $y$, the disk has radius $x=\frac{1}{y}$ and thickness $\mathrm{d}y$, so its volume is $\frac{\pi}{y^2}\mathrm{d}y$. The base of the solid is at height $y=1$, and its top is at height $y=\frac{1}{a}$. So, the volume of the entire solid is:

$$\begin{aligned} \int_1^{1/a} \frac{\pi}{y^2}\mathrm{d}y = \left[-\frac{\pi}{y}\right]_1^{1/a} = \pi (1-a) \end{aligned}$$

If we imagine sliding $a$ closer and closer to 0, the volume increases, getting closer and closer to $\pi$ units, but never quite reaching it.

So, the statement is false. For example, if we set $M=4$, no matter which $a$ we choose our solid has volume strictly less than $M$.

Remark: we've seen before that $\int_0^1 \frac{1}{x}\,\mathrm{d}x$ diverges. If we imagine the solid that would result from choosing $a=0$, it would have a scant volume of $\pi$ cubic units, but a silhouette (side view) of infinite area.

There are two sources of error: the upper bound is $t$, rather than infinity, and we're using an approximation with some finite number of intervals, $n$. Our plan is to first find a value of $t$ that introduces an error of no more than $\frac{1}{2}10^{-4}$. That is, we'll find a value of $t$ such that $\int_t^\infty \frac{e^{-x}}{1+x}~\mathrm{d}x \leq \frac{1}{2}10^{-4}$. After that, we'll find a value of $n$ that approximates $\int_0^t \frac{e^{-x}}{x+1}~\mathrm{d}x$ to within $\frac{1}{2}10^{-4}$. Then, all together, our error will be at most $\frac{1}{2}10^{-4} + \frac{1}{2}10^{-4} = 10^{-4}$, as desired. (Note we could have broken up the error in another way—it didn't have to be $\frac{1}{2}10^{-4}$ and $\frac{1}{2}10^{-4}$. This will give us one of many possible answers.)

Let's find a $t$ such that $\int_t^\infty \frac{e^{-x}}{1+x}\ \mathrm{d}x\le\frac{1}{2} 10^{-4}$. For all $x\ge 0$, $0<\frac{e^{-x}}{1+x}\le e^{-x}$, so

$$\begin{aligned} \int_t^\infty \frac{e^{-x}}{1+x}\ \mathrm{d}x \le \int_t^\infty e^{-x}\ \mathrm{d}x =e^{-t}&\stackrel{(*)}{\le}\frac{1}{2} 10^{-4}\\ \text{where ($*$) is true if }\quad t&\ge -\log\Big(\frac{1}{2} 10^{-4}\Big)\approx 9.90 \end{aligned}$$

Choose, for example, $t=10$.

Now it's time to decide how many intervals we're going to use to approximate $\displaystyle\int_0^t \frac{e^{-x}}{x+1}~\mathrm{d}x$. Again, we want our error to be less than $\frac{1}{2}10^{-4}$. To bound our error, we need to know the second derivative of $\frac{e^{-x}}{x+1}$.

$$\begin{aligned} f(x)=\frac{e^{-x}}{1+x} \implies f'(x)=-\frac{e^{-x}}{1+x}-\frac{e^{-x}}{(1+x)^2} \implies f''(x)&=\frac{e^{-x}}{1+x}+2\frac{e^{-x}}{(1+x)^2} +2\frac{e^{-x}}{(1+x)^3} \end{aligned}$$

Since $f''(x)$ is positive, and decreases as $x$ increases,

$$\begin{aligned} |f''(x)|\le f''(0) = 5 \implies |E_n|\le\frac{5(10-0)^3}{24n^2}=\frac{5000}{24n^2} =\frac{625}{3n^2} \end{aligned}$$

and $|E_n|\le \frac{1}{2} 10^{-4}$ if

$$\begin{aligned} &&\frac{625}{3n^2}&\le \frac{1}{2} 10^{-4}\\ &\iff\qquad & n^2&\ge \frac{1250\times 10^4}{3}\\ &\iff\quad &n&\ge \sqrt{\frac{1.25\times 10^7}{3}} \approx 2041.2 \end{aligned}$$

So $t=10$ and $n= 2042$ will do the job. There are many other correct answers.

Define $F(x)= \int_0^x \frac{1}{e^{t}}\mathrm{d}t$.

$$\begin{aligned} F(x)=\int_0^x \frac{1}{e^{t}}\mathrm{d}t&=\left[-\frac{1}{e^{t}}\right]_0^x = \frac{1}{e^0}-\frac{1}{e^{x}}<\frac{1}{e^0}=1 \end{aligned}$$

So, the statement is false: there is no $x$ such that $F(x)=1$. For every real $x$, $F(x) < \frac{1}{e^0}=1$.

We note here that $\displaystyle\lim_{x \to \infty} \int_0^x\frac{1}{e^{t}}\mathrm{d}t=1$. So, as $x$ grows larger, the gap between $F(x)$ and 1 grows infintesimally small. But there is no real value of $x$ where $F(x)$ is exactly equal to 1.