函数 f 定义为 $f(x)=\ln(5-x),\; x<5$。

1. Find the limits of integration along the $y$-axis.

The region is bounded by the $x$-axis and the $y$-axis.
On the $x$-axis, $y = 0$.
On the $y$-axis, $x = 0$, so $y = \ln(5 - 0) = \ln 5$.
The limits of integration for $y$ are from $0$ to $\ln 5$.

2. Express $x$ in terms of $y$.

$y = \ln(5 - x)$
$e^y = 5 - x$
$x = 5 - e^y$

3. Set up the integral for the volume of revolution about the $y$-axis.

The volume $V$ obtained by rotating about the $y$-axis is given by $\pi \int x^2 \, dy$.
$V = \pi \int_{0}^{\ln 5} (5 - e^y)^2 \, dy$
$V = \pi \int_{0}^{\ln 5} (25 - 10e^y + e^{2y}) \, dy$

4. Evaluate the integral.

$V = \pi \left[ 25y - 10e^y + \frac{1}{2}e^{2y} \right]_0^{\ln 5}$
$V = \pi \left( \left( 25\ln 5 - 10e^{\ln 5} + \frac{1}{2}e^{2\ln 5} \right) - \left( 0 - 10e^0 + \frac{1}{2}e^0 \right) \right)$
$V = \pi \left( \left( 25\ln 5 - 10(5) + \frac{1}{2}(25) \right) - \left( -10 + \frac{1}{2} \right) \right)$
$V = \pi (25\ln 5 - 50 + 12.5 + 9.5)$
$V = \pi (25\ln 5 - 28)$
$V \approx 38.440188$
$V \approx 38.44 \text{ (to 2 d.p.)}$

1. Use the property that the modulus of a function is non-negative to restrict the domain of $|x|$.

Since the right-hand side is $|\mathrm{f}(x)|$, which is non-negative, the left-hand side must also be non-negative:
$\mathrm{f}(|x|) \ge 0$
$\ln(5-|x|) \ge 0$
$5-|x| \ge 1 \implies |x| \le 4$
This restricts $x$ to the interval $-4 \le x \le 4$.

2. Simplify $|\mathrm{f}(x)|$ for the restricted domain and solve for $x$.

For $x$ in the interval $[-4, 4]$, we have $5-x \ge 1$, which means $\mathrm{f}(x) = \ln(5-x) \ge 0$.
Since $\mathrm{f}(x)$ is non-negative on this interval, $|\mathrm{f}(x)| = \mathrm{f}(x) = \ln(5-x)$.
Equating both sides of the original equation:
$\ln(5-|x|) = \ln(5-x)$
$5 - |x| = 5 - x \implies |x| = x$
This implies that $x \ge 0$.
Combining $x \ge 0$ with $|x| \le 4$, we obtain:
$0 \le x \le 4$

1. Reduce the combination of integrals to a single integral using (b)(i)

$f(x)=\ln(5-x)$ has its zero at $x=4$, so $f(x)\ge 0$ on $[0,4]$ and $f(x)<0$ on $(4,\tfrac92]$. On $[0,4]$ the two integrands coincide, so those parts cancel; on $(4,\tfrac92]$ they differ by $2f(x)$ (since $|f(x)|=-f(x)$ there). Hence
$\int_{0}^{9/2} f(x)\,dx-\int_{9/2}^{0} f(|x|)\,dx \;=\; 2\int_{4}^{9/2}\ln(5-x)\,dx.$

2. Integrate ln(5−x) by parts

$\int \ln(5-x)\,dx = x\ln(5-x)+\int \frac{x}{5-x}\,dx = x\ln(5-x)+\int\!\Big(-1+\frac{5}{5-x}\Big)dx$
$= x\ln(5-x)-x-5\ln(5-x)+C = (x-5)\ln(5-x)-x+C.$

3. Substitute the limits and obtain the final value

$2\Big[(x-5)\ln(5-x)-x\Big]_{4}^{9/2}$
At $x=\tfrac92$: $(-\tfrac12)\ln\tfrac12-\tfrac92=\tfrac12\ln2-\tfrac92$. At $x=4$: $(-1)\ln1-4=-4$.
$=2\Big[\big(\tfrac12\ln2-\tfrac92\big)-(-4)\Big]=2\big(\tfrac12\ln2-\tfrac12\big)=\ln2-1.$