91Ó°ÊÓ

The area under a curve in calculus refers to the region bounded by the curve, the x-axis, and vertical lines corresponding to specific values of x. Calculating the area gives us insights into the space occupied by the region. In the given exercise, the region is bounded by the graph of the function \( y = \ln x \), the x-axis \( y = 0 \), and the vertical line \( x = e \). To find the area, we set up an integral that computes the difference between the bounding function above the x-axis and the x-axis itself. The integral is:\[ \int_{1}^{e} \ln x \, dx \]Evaluate this integral to find the area:- Integrate \( \ln x \): the anti-derivative is \( x \ln x - x + C \).- Apply the limits from 1 to \( e \): - Evaluate at \( x = e \): \( e \ln e - e = e - e = 0 \). - Evaluate at \( x = 1 \): \( 1 \ln 1 - 1 = -1 \).- Subtract to find the area: 0 - (-1) = 1.Thus, the area under the curve from \( x = 1 \) to \( x = e \) is 1 square unit.

Volumes of revolution involve rotating a given region around an axis to form a 3-dimensional solid. In this exercise, we revolve the region around both the x-axis and the line \( x = -2 \), requiring use of different techniques to find volumes.When rotating around the x-axis, the **disk method** is used:- Formula: \( V = \pi \int_{1}^{e} (\ln x)^2 \, dx \).- Here, each small disk has a radius determined by \( \ln x \) and thickness \( dx \).- Using integration by parts and simplifying, the evaluated volume is \( \pi e \) cubic units.For rotation around the line \( x = -2 \), the **washer method** is more appropriate:- Formula: \( V = \pi \int_{1}^{e} ((\ln x + 2)^2 - 4) \, dx \).- Each washer’s outer radius is \( \ln x + 2 \) and inner radius is 2.- Evaluation gives a volume of \( \frac{9\pi e}{4} \) cubic units.Understanding these methods helps in visualizing how areas revolve to create 3D shapes.

A centroid is the geometric center of a plane region, and finding it involves calculating both the x and y coordinates. This provides the balance point of the shape, like how you would balance a physical object.To find the x-coordinate (\( \bar{x} \)):- Formula: \( \bar{x} = \frac{1}{A} \int_{1}^{e} x(\ln x) \, dx \).- With the area \( A = 1 \), after evaluating: - Integral becomes \( \frac{e^2}{4} \). - Thus, \( \bar{x} = \frac{e}{4} \).The y-coordinate (\( \bar{y} \)) is found using:- Formula: \( \bar{y} = \frac{1}{A} \int_{1}^{e} \frac{1}{2} ( \ln x)^2 \, dx \).- After evaluation: - Integral results in \( \frac{1}{e} \). - Hence, \( \bar{y} = \frac{1}{e} \).Finding the centroid is vital for engineering and physics applications, marking where symmetry and balance occur.

Integration techniques are essential tools in calculus for tackling various integrals. Different techniques suit different problems, enhancing our capacity to solve otherwise complicated integrals.In the area and volume calculations, **integration by parts** was crucial. This technique stems from the product rule of differentiation and is used when dealing with products of functions:- For volumes, especially around the x-axis, integrating \( (\ln x)^2 \) required integration by parts multiple times to simplify.- **L-iteration** can be used: choose each time a simpler term for the function that will be reduced through differentiation.Each integral, though it might appear difficult, can often be simplified into basic steps that yield thorough results. Mastering these techniques empowers you to solve numerous calculus problems with confidence.