91Ó°ÊÓ

Use a graphing utility, such as Desmos or GeoGebra, to graph the function \(f(x) = x + \sin(2x)\) for \(x \in [0, \pi]\). By looking at the graph, we can better understand the shape of the solid that will be generated when revolving the function about the x-axis.

Now, use a computer algebra system (CAS) to find the volume of the solid formed by revolving the region between the graph of \(f(x)\) and the x-axis around the x-axis. Most CAS will have an option to evaluate integrals. The formula for the volume of a solid of revolution generated by revolving a function about the x-axis is: \[V = \pi \int_{a}^{b} [f(x)]^2 dx\] Here, we have \(a = 0\), \(b = \pi\), and \(f(x) = x + \sin(2x)\). Plug these values into the formula and use the CAS to compute the integral: \[V = \pi \int_{0}^{\pi} (x+\sin(2x))^2 dx\]

Now, let's calculate the volume exactly by carrying out the integration. We start with the formula for the volume: \[V = \pi \int_{0}^{\pi} (x + \sin(2x))^2 dx\] First, expand the square inside the integral: \[(x + \sin(2x))^2 = x^2 + 2x \sin(2x) + (\sin(2x))^2\] Now, substitute this back into the integral: \[V = \pi \int_{0}^{\pi} \left(x^2 + 2x \sin(2x) + (\sin(2x))^2\right) dx\] Next, we'll break this integral into three separate integrals: \[V = \pi\left[\int_{0}^{\pi} x^2 dx + \int_{0}^{\pi} 2x \sin(2x) dx + \int_{0}^{\pi} (\sin(2x))^2 dx\right]\] Now, let's evaluate each integral separately: (1) \(\int_{0}^{\pi} x^2 dx = \dfrac{x^3}{3}\Big|_0^\pi = \dfrac{\pi^3}{3}\) (2) Use integration by parts for the second integral: Let \(u = x\), \(dv = 2x\sin(2x)dx\), so we have \(du = dx\), and \(v = -\dfrac{1}{2}\cos(2x)\). \(\int_{0}^{\pi} 2x\sin(2x) dx = \left[-x\frac{1}{2}\cos(2x)\right]_0^\pi + \int_0^\pi \frac{1}{2}\cos(2x) dx\) \(\int_{0}^{\pi} 2x\sin(2x) dx = -\dfrac{1}{2}\pi\cos(2\pi) + 0 +\left[ \frac{1}{4}\sin(2x)\right]_0^\pi\) \(\int_{0}^{\pi} 2x\sin(2x) dx = -\dfrac{1}{2} \pi\) (3) Use double-angle identity \(\sin^2(2x) = \dfrac{1 - \cos(4x)}{2}\) for the third integral: \(\int_{0}^{\pi} \sin^2(2x) dx = \dfrac{1}{2}\int_{0}^{\pi} (1-\cos(4x)) dx\) \(\int_{0}^{\pi} \sin^2(2x) dx = \dfrac{1}{2}\left[x - \frac{1}{4} \sin(4x)\right]_0^\pi = \dfrac{1}{2}\pi\) Now, let's substitute the results of the three integrals back into our formula for the volume: \[V = \pi\left[\dfrac{\pi^3}{3} - \dfrac{1}{2} \pi + \dfrac{1}{2}\pi\right] = \dfrac{\pi^4}{3}\] The volume of the resulting solid is \(\dfrac{\pi^4}{3}\).