The problem asks us to find the area of the region $S$ bounded by the curve $y = \tan x$, the x-axis, and the line $x = \frac{\pi}{4}$ for $0 \le x \le \frac{\pi}{4}$. Then, we need to find the volume of the solid generated by rotating $S$ about the x-axis, given that $(\tan x - x)' = \tan^3 x$.

1. Problem Description

The problem asks us to find the area of the region $S$ bounded by the curve $y = \tan x$, the x-axis, and the line $x = \frac{\pi}{4}$ for $0 \le x \le \frac{\pi}{4}$. Then, we need to find the volume of the solid generated by rotating $S$ about the x-axis, given that $(\tan x - x)' = \tan^3 x$.

2. Solution Steps

(1) The area of the region $S$ is given by the integral of the function $y = \tan x$ from $0$ to $\frac{\pi}{4}$:

$A = \int_0^{\frac{\pi}{4}} \tan x dx$

Since $\tan x = \frac{\sin x}{\cos x}$, we can write this as:

$A = \int_0^{\frac{\pi}{4}} \frac{\sin x}{\cos x} dx$

Let $u = \cos x$, then $du = -\sin x dx$. When $x = 0$, $u = \cos 0 = 1$. When $x = \frac{\pi}{4}$, $u = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.

So, $A = \int_1^{\frac{\sqrt{2}}{2}} -\frac{1}{u} du = -\ln |u| \Big|_1^{\frac{\sqrt{2}}{2}} = -(\ln \frac{\sqrt{2}}{2} - \ln 1) = -(\ln \sqrt{2} - \ln 2 - 0) = -(\frac{1}{2} \ln 2 - \ln 2) = \frac{1}{2} \ln 2$.

Therefore, the area $S$ is $\frac{1}{2} \log 2$ (assuming log refers to the natural logarithm). Thus the box 1 is

2.

(2) The volume of the solid generated by rotating $S$ about the x-axis is given by:

$V = \pi \int_0^{\frac{\pi}{4}} (\tan x)^2 dx = \pi \int_0^{\frac{\pi}{4}} \tan^2 x dx$

We are given that $(\tan x - x)' = \tan^3 x$. We can also rewrite $\tan^2 x$ as $\sec^2 x - 1$. Thus,

$\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x$.

So, $V = \pi \int_0^{\frac{\pi}{4}} \tan^2 x dx = \pi [\tan x - x]_0^{\frac{\pi}{4}} = \pi [(\tan \frac{\pi}{4} - \frac{\pi}{4}) - (\tan 0 - 0)] = \pi [(1 - \frac{\pi}{4}) - (0 - 0)] = \pi (1 - \frac{\pi}{4}) = \pi (\frac{4 - \pi}{4})$

Thus, we are told $(\tan x - x)' = \tan^2(x)\sec^2(x) = \sec^2(x) - 1$ and not $\tan^3(x)$. So the box 3 is

2. So, volume is $V = \pi \int_0^{\pi/4} \tan^2(x) dx = \pi (\tan(\frac{\pi}{4}) - \frac{\pi}{4}) = \pi(1 - \frac{\pi}{4}) = \frac{\pi(4 - \pi)}{4}$. So, box 5 is

4.

3. Final Answer

(1) 2

(2) 2

(3) 4