Answer: The centroid of the region under \(y=x^{3/2}\) from \(x=1\) to \(x=4\) is

\(\left(\bar x,\bar y\right)=\left(\frac{635}{217},\frac{1275}{496}\right).\)

The arc length from \(x=5\) to \(x=28\) is \(139\).

The region is bounded by \(y=x^{3/2}\), the \(x\)-axis and the vertical lines \(x=1\) and \(x=4\).

Its area is

\(A=\int_1^4x^{3/2}\,\mathrm dx=\left[\frac25x^{5/2}\right]_1^4=\frac25(32-1)=\frac{62}{5}.\)

The \(x\)-coordinate of the centroid is

\(\bar x=\frac{1}{A}\int_1^4x\cdot x^{3/2}\,\mathrm dx=\frac{1}{A}\int_1^4x^{5/2}\,\mathrm dx.\)

Thus

\(\int_1^4x^{5/2}\,\mathrm dx=\left[\frac27x^{7/2}\right]_1^4=\frac27(128-1)=\frac{254}{7}.\)

So

\(\bar x=\frac{254/7}{62/5}=\frac{635}{217}.\)

For the \(y\)-coordinate of the centroid of the area under a curve,

\(\bar y=\frac{1}{A}\cdot\frac12\int_1^4y^2\,\mathrm dx.\)

Since \(y^2=x^3\),

\(\frac12\int_1^4y^2\,\mathrm dx=\frac12\int_1^4x^3\,\mathrm dx=\frac12\left[\frac14x^4\right]_1^4=\frac{255}{8}.\)

Hence

\(\bar y=\frac{255/8}{62/5}=\frac{1275}{496}.\)

Now find the arc length from \(x=5\) to \(x=28\). Since

\(y=x^{3/2},\qquad \frac{\mathrm dy}{\mathrm dx}=\frac32x^{1/2},\)

the arc length is

\(s=\int_5^{28}\sqrt{1+\left(\frac32\sqrt x\right)^2}\,\mathrm dx=\int_5^{28}\sqrt{1+\frac94x}\,\mathrm dx.\)

So

\(s=\frac12\int_5^{28}\sqrt{9x+4}\,\mathrm dx.\)

Let \(u=9x+4\). Then \(\mathrm du=9\,\mathrm dx\), and

\(s=\frac1{18}\int_{49}^{256}u^{1/2}\,\mathrm du=\frac1{27}\left[u^{3/2}\right]_{49}^{256}.\)

Therefore

\(s=\frac1{27}(256^{3/2}-49^{3/2})=\frac1{27}(4096-343)=139.\)