91Ó°ÊÓ

Integration by parts is a technique used to integrate products of functions. It simplifies an integral into a potentially easier one. The formula is derived from the product rule of differentiation, and it's expressed as:\[\int u \, dv = uv - \int v \, du\]In this specific problem, we used integration by parts to find the area of the region under the curve \( y = \ln x \) from \( x = 1 \) to \( x = e \). Here, we chose \( u = \ln x \) and \( dv = dx \). It's useful to remember that this method is about choosing \( u \) and \( dv \) such that the resulting \( du \) and \( v \) make the integral simpler to solve.

• Differentiate \( u \) to find \( du \).
• Integrate \( dv \) to find \( v \).

After applying the integration by parts and calculating the definite integral, we get the area \( A = e - 1 \). This approach is particularly useful when dealing with logarithms or polynomial functions within integrals.

The disk method is a way to find the volume of a solid of revolution. This technique is typically applied when the solid is generated by revolving a region around the x-axis. It involves slicing the solid perpendicular to the axis of revolution into thin disks, then summing the volumes of these disks.For each small disk slice, the volume is defined as \( \pi \times (\text{radius})^2 \times \text{thickness} \). Here, the "thickness" is an infinitesimally small change in the x-direction, denoted as \( dx \), and the "radius" corresponds to the function value, \( y = \ln x \).The formula for the volume \( V \) using the disk method is:\[V = \pi \int_{a}^{b} (f(x))^2 \, dx\]In the problem, the function \( f(x) = \ln x \) is squared within the integral. The limits, from \( 1 \) to \( e \), define the bounds of the area being revolved. Calculating this integral yielded the volume of the solid around the x-axis.

The shell method is another technique to find the volume of a solid of revolution. It is particularly useful when revolving around a vertical line, such as \( x = -2 \) in this problem. Instead of disks, you visualize the solid as made of cylindrical shells.The volume of each shell is calculated based on its circumference, height, and thickness, with the formula:\[V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx\]For our scenario, the radius of each shell is \( x + 2 \), and the height is given by \( \ln x \). By integrating from \( x = 1 \) to \( x = e \), we sum these cylindrical volumes. This produces the volume of the solid formed by rotating the region around the line \( x = -2 \), resulting in a calculated volume of \( \frac{3\pi}{2} (e - 1) \).

The centroid of a planar region is the geometric center, also known as the center of mass or "balance point." This concept applies not only to physical objects but also to areas defined by mathematical curves.To find the centroid \((\bar{x}, \bar{y})\) of a region defined by a curve from \( x = a \) to \( x = b \), you will need:- The area \( A \) of the region. - Formulas for \( \bar{x} \) and \( \bar{y} \):\[\bar{x} = \frac{1}{A} \int_{a}^{b} x \, f(x) \, dx\]\[\bar{y} = \frac{1}{2A} \int_{a}^{b} (f(x))^2 \, dx\]In the given task, \( f(x) = \ln x \), the limits are \( x = 1 \) to \( x = e \), and the previously calculated area \( A = e - 1 \). Solving these integrals gives the coordinates of the centroid, \( \bar{x} = \frac{e+1}{2} \) and \( \bar{y} = \frac{\ln e - \ln 2}{2(e - 1)} \), providing a complete understanding of the centroid's position.