求 $\displaystyle \int \frac{9u-8}{4+9u^2}\,\mathrm{d}u$。 曲线 $C$ 由参数方程给出： $x = u^2 + u + 1$，$y = \dfrac{9u}{4+9u^2}$，其中 $u \ge 0$。 求由 $C$、x 轴以及直线 $x = 3$ 所围成区域的面积的精确值。 求当由直线 $x = -4$、$y = 2$ 以及椭圆 $(x+2)^2 + 4(y-1)^2 = 4$ 所围成的阴影区域绕 y 轴旋转 $2\pi$ 弧度时所形成的几何体的体积。结果保留一位小数。

1. Split the fraction into two separate terms to apply standard integration rules.

$\int \frac{9u-8}{4+9u^2}\,\mathrm{d}u = \int \frac{9u}{4+9u^2}\,\mathrm{d}u - \int \frac{8}{4+9u^2}\,\mathrm{d}u$

2. Integrate the first term using the form $\int \frac{f'(u)}{f(u)}\,\mathrm{d}u$.

$\int \frac{9u}{4+9u^2}\,\mathrm{d}u = \frac{1}{2} \int \frac{18u}{4+9u^2}\,\mathrm{d}u = \frac{1}{2} \ln(4+9u^2)$

3. Integrate the second term using the standard integral $\int \frac{1}{a^2+x^2}\,\mathrm{d}x = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)$.

$\int \frac{8}{4+9u^2}\,\mathrm{d}u = 8 \int \frac{1}{2^2+(3u)^2}\,\mathrm{d}u = 8 \left( \frac{1}{3} \right) \left( \frac{1}{2} \right) \tan^{-1}\left(\frac{3u}{2}\right) = \frac{4}{3} \tan^{-1}\left(\frac{3u}{2}\right)$

4. Combine the evaluated integrals and include the constant of integration.

$\frac{1}{2} \ln(4+9u^2) - \frac{4}{3} \tan^{-1}\left(\frac{3u}{2}\right) + C$

1. Evaluate the integral from part (a)(i) as it is required for this part.

$\int \frac{9u-8}{4+9u^2} \, \mathrm{d}u = \int \frac{9u}{4+9u^2} \, \mathrm{d}u - \int \frac{8}{4+9u^2} \, \mathrm{d}u$
$= \frac{1}{2} \ln(4+9u^2) - \frac{8}{3}\left(\frac{1}{2}\right)\arctan\left(\frac{3u}{2}\right) + C$
$= \frac{1}{2} \ln(4+9u^2) - \frac{4}{3} \arctan\left(\frac{3u}{2}\right) + C$

2. Find the values of the parameter $u$ corresponding to the integration limits.

The area is bounded by the x-axis ($y = 0$) and the line $x = 3$.
When $y = 0$:
$\frac{9u}{4+9u^2} = 0 \implies u = 0$
When $x = 3$:
$u^2 + u + 1 = 3 \implies u^2 + u - 2 = 0$
$(u+2)(u-1) = 0$
Since $u \ge 0$, $u = 1$.
Thus, the limits of integration for $u$ are from $0$ to $1$.

3. Set up the definite integral for the area using the parametric equations.

The formula for the area under a parametric curve is $\int y \frac{\mathrm{d}x}{\mathrm{d}u} \, \mathrm{d}u$.
$\frac{\mathrm{d}x}{\mathrm{d}u} = 2u + 1$
$\text{Area} = \int_0^1 \left( \frac{9u}{4+9u^2} \right) (2u + 1) \, \mathrm{d}u$
$= \int_0^1 \frac{18u^2 + 9u}{4+9u^2} \, \mathrm{d}u$

4. Rewrite the integrand to express it in terms of the integral from part (a)(i).

By performing algebraic manipulation (such as long division):
$\frac{18u^2 + 9u}{4+9u^2} = \frac{2(9u^2 + 4) + 9u - 8}{4+9u^2}$
$= 2 + \frac{9u - 8}{4+9u^2}$

5. Evaluate the definite integral using the pre-calculated result to find the exact area.

$\text{Area} = \int_0^1 \left( 2 + \frac{9u - 8}{4+9u^2} \right) \, \mathrm{d}u$
$= \left[ 2u + \frac{1}{2} \ln(4+9u^2) - \frac{4}{3} \arctan\left(\frac{3u}{2}\right) \right]_0^1$
Substituting upper limit $u = 1$:
$= \left( 2 + \frac{1}{2}\ln 13 - \frac{4}{3}\arctan\left(\frac{3}{2}\right) \right)$
Substituting lower limit $u = 0$:
$= \left( 0 + \frac{1}{2}\ln 4 - 0 \right) = \ln 2$
Subtracting the values:
$\text{Area} = 2 + \frac{1}{2}\ln 13 - \frac{4}{3}\arctan\left(\frac{3}{2}\right) - \ln 2$
$= 2 + \frac{1}{2}\ln\left(\frac{13}{4}\right) - \frac{4}{3}\arctan\left(\frac{3}{2}\right)$

1. Express the x-coordinate of the left portion of the ellipse in terms of $y$.

From the ellipse equation $(x+2)^2 + 4(y-1)^2 = 4$, we solve for $x$:
$(x+2)^2 = 4 - 4(y-1)^2$
$x + 2 = \pm \sqrt{4 - 4(y-1)^2}$
Since the shaded region is bounded by the left side of the ellipse where $x < -2$, we take the negative root:
$x = -2 - \sqrt{4 - 4(y-1)^2}$

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

The solid is formed by rotating the region between the vertical line $x = -4$ and the ellipse about the y-axis, for $1 \le y \le 2$.
The distance from the y-axis to the outer boundary is $R = |-4| = 4$.
The distance from the y-axis to the inner boundary is $r = |x|$.
The volume of revolution using the washer method is:
$V = \pi \int_{1}^{2} \left( R^2 - r^2 \right) \,\mathrm{d}y$
$V = \pi \int_{1}^{2} \left[ 4^2 - x^2 \right] \,\mathrm{d}y$
$V = \pi \int_{1}^{2} \left[ 16 - \left( -2 - \sqrt{4 - 4(y-1)^2} \right)^2 \right] \,\mathrm{d}y$

3. Evaluate the integral using a graphing calculator.

Using a graphing calculator to evaluate the definite integral directly:
$V \approx 9.5823$
$V = 9.6 \text{ (to 1 d.p.)}$