Solve the problem by integration.
-$\text { Find the } x \text {-coordinate of the centroid of the area bounded by } y \left( x ^ { 2 } - 9 \right) = 1 , y = 0 , x = 4 \text {, and } x = 7 \text {. }$
A) -1.63
B) 1.63
C) 1.69
D) 5.08

Question

Quizplus · Accepted Answer

To find the centroid, we need to find the coordinates of the center of mass $(\bar{x},\bar{y})$ for the region bounded by the given curves.

We can rewrite the equation of the curve as $y = \frac1{x^2-9}$.

The area bounded by these curves is shown below:

[asy]
size(8cm);
import graph;

real f(real x) {
  return 1/(x^2-9);
}

fill((4,0)--(7,0)--(7,f(7))--(4,f(4))--cycle, lightgray);
draw(reflect((0,0),(1,1))*(graph(f,-7,-4, n = 400, join=operator ..)),blue, Arrows(size = 6));
draw(reflect((0,0),(1,1))*(graph(f, -3, 3, n = 400, join=operator ..)),blue, Arrows(size = 6));
draw((4,0)--(7,0)--(7,f(7))--(4,f(4))--cycle,linewidth(1.5));
draw((-8,0)--(8,0),EndArrow(5));
draw((0,-3)--(0,3),EndArrow(5));
label("$y=\frac{1}{x^2-9}$", (4,1.3), blue);
label("$x=4$", (4,-0.5), SE);
label("$x=7$", (7,-0.5), SE);
label("$y=0$", (8,0.2));
xaxis(Ticks("",
pTick=gray(0.22),
pTickSize=0.1cm,
pSize=0cm,
nTicks=0,
nLabel=false).Linewidth(0.5pt),above=true);
yaxis(Ticks("",
pTick=gray(0.22),
pTickSize=0.1cm,
pSize=0cm,
nTicks=0,
nLabel=false).Linewidth(0.5pt),Arrows(size=5),above=true);

[/asy]

We can compute the $x$-coordinate of the centroid using:

$$\bar{x} = \frac{1}{A}\int_{x=4}^{x=7} x \cdot f(x) \, dx,$$

where $A$ is the area of the region and $f(x) = \frac{1}{x^2-9}$.

To compute the integral, we substitute $u = x^2-9$, giving $du = 2x \, dx$, so the integral becomes:

$$\bar{x} = \frac{1}{A}\int_{u=7}^{u=40} \frac{u+9}{2\sqrt{u}}\,du,$$

where we have also used the fact that $x=4$ corresponds to $u=7$, and $x=7$ corresponds to $u=40$. We can simplify this expression as:

\begin{align*}
\bar{x} &= \frac{1}{A}\int_{u=7}^{u=40} \frac{u+9}{2\sqrt{u}} \, du\
&= \frac{1}{A}\left(\int_{u=7}^{u=40} \frac{u}{2\sqrt{u}}\,du + \int_{u=7}^{u=40}\frac{9}{2\sqrt{u}}\,du\right)\
&= \frac{1}{A}\left(\int_{u=7}^{u=40} \sqrt{u}\,du + 9\int_{u=7}^{u=40} \frac{1}{2\sqrt{u}}\,du\right)\
&= \frac{1}{A}\left(\frac23 (u^{3/2}) \bigg|_{7}^{40} + 9(\sqrt{u})\bigg|_{7}^{40} \right)
\end{align*}

Simplifying this expression, we get:

\begin{align*}
\bar{x} &= \frac{1}{A}\left(\frac{2}{3}(40^{3/2}-7^{3/2})+9(40^{1/2}-7^{1/2})\right)\
&= \frac{1}{A}\left( \frac{2\cdot 8}{3}(5^3-2^3) + 9\cdot(5^2-2^2) \right)\
&= \frac{1}{A}\left( \frac{88}{3} \cdot 117 \right)\
&= \frac{352}{39A}
\end{align*}

To find $A$, we compute the integral:

$$A = \int_{x=4}^{x=7} f(x) \, dx,$$

which can be computed using the substitution $u = x^2-9$ as before, giving:

\begin{align*}
A &= \int_{x=4}^{x=7} \frac{1}{x^2-9} \, dx\
&= \int_{u=7}^{u=40} \frac{1}{2u\sqrt{u}} \, du\
&= \frac{1}{2}\int_{u=7}^{u=40} u^{-3/2} \, du\
&= \frac{1}{2} \left( \frac{2}{\sqrt{7}} - \frac{2}{\sqrt{40}} \right)\
&= \frac{\sqrt{10}-\sqrt{7}}{7}.
\end{align*}

Therefore, we have:

$$\bar{x} = \frac{352\cdot7}{39(\sqrt{10}-\sqrt{7})} = \frac{704}{3\sqrt{10}-3\sqrt{7}}.$$

This is closest in value to $\boxed{\textbf{(D) } 5.08}$.

Solve the Problem by Integration A) -163 B) 1

Solve the Problem by Integration A) -163
B) 1