이상적분

backwards

$5\sin\theta$

(a) $\tan^2\theta + 1 = \sec^2\theta$ (b) $9\sec^2\theta$.

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.

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.