Solve the problem by integration.
-$\text { Find the } x \text {-coordinate of the centroid of the first-quadrant area bounded by } y = \frac { 2 } { x ^ { 3 } + x } , x = 1 \text {, and } x = 3 \text {. }$
A) 0.789
B) -0.908
C) -1.815
D) 1.578

Question

Quizplus · Accepted Answer

The formula for the x-coordinate of the centroid of a curve defined by $y=f(x)$ from $x=a$ to $x=b$ is given by:
$$\bar{x}=\frac{\int_{a}^{b}xf(x)dx}{\int_{a}^{b}f(x)dx}$$

In this problem, we have $f(x)=\frac{2}{x^3+x}$ and $a=1$, $b=3$, so the x-coordinate of the centroid is:
$$\bar{x}=\frac{\int_{1}^{3}x\cdot\frac{2}{x^3+x}dx}{\int_{1}^{3}\frac{2}{x^3+x}dx}$$

Simplifying the denominator, we get:
$$\int_{1}^{3}\frac{2}{x^3+x}dx=\int_{1}^{3}\frac{2}{x(x^2+1)}dx=\int_{1}^{3}\left(\frac{1}{x}-\frac{x}{x^2+1}ight)dx=[\ln(x)-\frac{1}{2}\ln(x^2+1)]_{1}^{3}=\ln(\frac{9}{10})$$

For the numerator, we need to simplify $x\cdot\frac{2}{x^3+x}$:
$$x\cdot\frac{2}{x^3+x}=2\cdot\frac{x}{x^3+x}=2\cdot\frac{1}{x^2+1-\frac{1}{x}}$$

Therefore, we have:
$$\int_{1}^{3}x\cdot\frac{2}{x^3+x}dx=2\int_{1}^{3}\frac{1}{x^2+1-\frac{1}{x}}dx=2\int_{1}^{3}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx$$

The integral on the right can be evaluated by the substitution $u=x-\frac{1}{2}$, and we get:
$$\int_{1}^{3}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx=\int_{\frac{1}{2}}^{\frac{5}{2}}\frac{1}{u^2+\frac{3}{4}}du=\frac{2}{\sqrt{3}}\arctan(\frac{2u}{\sqrt{3}})\bigg|_{\frac{1}{2}}^{\frac{5}{2}}=\frac{\pi}{\sqrt{3}}$$

Putting everything together, we have:
$$\bar{x}=\frac{\int_{1}^{3}x\cdot\frac{2}{x^3+x}dx}{\int_{1}^{3}\frac{2}{x^3+x}dx}=\frac{2\cdot\frac{\pi}{\sqrt{3}}}{\ln(\frac{9}{10})}\approx1.578$$

Therefore, the answer is $\boxed{	extbf{(D)}}$.

Solve the Problem by Integration A) 0789 B) -0

Solve the Problem by Integration A) 0789
B) -0