一曲线的斜率函数为 $\frac{dy}{dx} = -\frac{1}{x^2} + 6 \cos 3x$。曲线上的点 $P$ 的坐标为 $\left(\frac{\pi}{2}, \pi\right)$。存在一点 $Q$ 使得 $PQ$ 的中点为 $\left(\frac{3}{4}\pi, 2\pi\right)$。求

1. Set up and solve the equation for the $x$-coordinate of $Q$

Let the coordinates of $Q$ be $(x, y)$.
Using the midpoint formula for the $x$-coordinate:
$\frac{\frac{\pi}{2} + x}{2} = \frac{3}{4}\pi$
$\frac{\pi}{2} + x = \frac{3}{2}\pi$
$x = \frac{3}{2}\pi - \frac{\pi}{2} = \pi$

2. Set up and solve the equation for the $y$-coordinate of $Q$

Using the midpoint formula for the $y$-coordinate:
$\frac{\pi + y}{2} = 2\pi$
$\pi + y = 4\pi$
$y = 3\pi$

1. Integrate the gradient function to find the general equation of the curve

$\frac{dy}{dx} = -x^{-2} + 6 \cos 3x$
$y = \int (-x^{-2} + 6 \cos 3x) \, dx$
$y = -\frac{x^{-1}}{-1} + \frac{6 \sin 3x}{3} + c$
$y = \frac{1}{x} + 2 \sin 3x + c$

2. Substitute the coordinates of point P to find the constant of integration $c$

Substitute $x = \frac{\pi}{2}$ and $y = \pi$:
$\pi = \frac{1}{\frac{\pi}{2}} + 2 \sin\left(3\left(\frac{\pi}{2}\right)\right) + c$
$\pi = \frac{2}{\pi} + 2 \sin\left(\frac{3\pi}{2}\right) + c$
$\pi = \frac{2}{\pi} + 2(-1) + c$
$c = \pi - \frac{2}{\pi} + 2$

3. State the final equation of the curve

$y = \frac{1}{x} + 2 \sin 3x + \pi - \frac{2}{\pi} + 2$