Applications of Integration

Part 1: Finding the Original Equation of the Curve

$\begin{aligned}
 \frac{\text{d}y}{\text{d}x}&=\int{18{{\left( x-2 \right)}^{-3}}\text{d}x} \\
& =\frac{18{{\left( x-2 \right)}^{-2}}}{\left( -2 \right)}+c \\
 \frac{\text{d}y}{\text{d}x}&=-9{{\left( x-2 \right)}^{-2}}+c
\end{aligned}$

Solve for the $c$

When $\frac{\text{d}y}{\text{d}x}=2,x=5$:
$\begin{aligned}
 2&=-9{{\left( 5-2 \right)}^{-2}}+c \\
 c&=2+9{{\left( 3 \right)}^{-2}} \\
& =3
\end{aligned}$

$\frac{\text{d}y}{\text{d}x}=-9{{\left( x-2 \right)}^{-2}}+3$

Part 2: Finding the Equation of the Line

$\begin{aligned}
 y&=\int{-9{{\left( x-2 \right)}^{-2}}+3\,\text{d}x} \\
 y&=\frac{-9{{\left( x-2 \right)}^{-1}}}{\left( -1 \right)}+3x+D \\
 y&=\frac{9}{x-2}+3x+D
\end{aligned}$

Substitute coordinates into the equation
When $x=5,y=20$:
$\begin{aligned}
 20&=\frac{9}{5-2}+3\left( 5 \right)+D \\
 D&=2
\end{aligned}$

Equation of curve $y=\frac{9}{x-2}+3x+2$

Part 3: Find the $y$-intercept, when $x=0$

$\begin{aligned}
 y&=\frac{9}{0-2}+3\left( 0 \right)+2 \\
 y&=-\frac{5}{2}
\end{aligned}$

$\left( 0,-\frac{5}{2} \right)$

Find the gradient of the tangent

$\begin{aligned}
 {{m}_{\operatorname{t}\text{angent}}},{{\left. \frac{\text{d}y}{\text{d}x} \right|}_{x=0}}&=-9{{\left( 0-2 \right)}^{-2}}+3 \\
& =\frac{3}{4}
\end{aligned}$

Find the equation of the tangent

$\begin{aligned}
 y-\left( -\frac{5}{2} \right)&=\frac{3}{4}\left( x-0 \right) \\
 y&=\frac{3}{4}x-\frac{5}{2}
\end{aligned}$