Question

Find the area of the region bounded by the graphs of the given equations. $$ \begin{aligned} &y=x \ln x\\\ &y=0, x=e \end{aligned} $$

Short answer

The area of the region bounded by the given curves is \(e - \frac{1}{4}e^2 + \frac{1}{4}\) square units.

Step-by-step solution

Setting up the integral

First, the definite integral needs to be set up. Since the boundaries are \(x=e\) and \(y=0\), we know the area will be between these points. The function given is \(f(x) = x \ln x\), thus the definite integral to calculate is: \(\int_{1}^{e} x \ln x \, dx\).

Solving the integral

The integral \(\int_{1}^{e} x \ln x \, dx\) is not an easy one to solve. However, by using integration by parts, with \(u=\ln x\) and \(dv=x dx\), it can be solved. After deriving and integrating accordingly to get \(du = \frac{1}{x} dx\) and \(v= \frac{1}{2}x^2\), the integral \(\int_{1}^{e} x \ln x \, dx\) will become \(\int_{1}^{e} u dv\). Apply the integration by parts formula: \(\int u dv = uv - \int v du\), this gives: \(x \ln x\Big|_1^e - \int_{1}^{e} \frac{1}{2}x \, dx\). This integral can now be computed to calculate the area under the curve.

Evaluating the integral

The definite integral can now be solved. Evaluating \(x \ln x\Big|_1^e\) gives \(e * \ln e - 1 * \ln 1 = e - 0 = e\).\nThen, evaluating \(- \int_{1}^{e} \frac{1}{2}x \, dx\) gives \(- \frac{1}{4}x^2\Big|_1^e = - \frac{1}{4}e^2 + \frac{1}{4}\). Adding these results together gives the final value for the area under the curve, which is \(e - \frac{1}{4}e^2 + \frac{1}{4}\).

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus. It's used to calculate the exact area under a curve within specified boundaries. When you see an integral written as \( \int_a^b f(x) \, dx \), the values \( a \) and \( b \) are the limits that show where to start and stop on the x-axis. This encapsulates the area between the curve of \( f(x) \) and the x-axis from \( x = a \) to \( x = b \).

Think of definite integrals as painting the area under a curve between two x-values. Every tiny slice or rectangle under the curve adds up to form this area.

Here's how it works: you slice the region into infinitesimally small rectangles, calculate the sum of the areas of these tiny slices, and take the limit as the slice size approaches zero.

• The top of each rectangle follows the curve.
• The base of each rectangle lies on the x-axis between the limits \( a \) and \( b \).

In this exercise, the definite integral is set from \( x = 1 \) to \( x = e \).

Area under Curve

Finding the area under a curve is a powerful tool for understanding a wide variety of real-world phenomena. This process helps determine the space covered or the total value of something represented by a curve.

In the context of \( y = x \ln x \) between \( x=1 \) and \( x=e \), calculating this specific area under the curve can tell us about the behavior and properties of the function over this interval.

It's like plotting how much space is occupied under this specific functional expression, which visually represents the accumulation of value over the specified range. To compute it accurately, we use techniques such as integration by parts.

Integration by parts is a technique derived from the product rule of differentiation. It's used when integrating products of functions where simple techniques don't work. In our example, this allows us to solve the integral \( \int x \ln x \, dx \).

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a logarithm to the base \( e \), where \( e \) is the constant approximately equal to 2.71828. It's a fundamental mathematical concept that shows up frequently in calculus and mathematical modeling.

This logarithm is particularly useful in integration and differentiation since it simplifies the relationship and operations involving exponential and logarithmic functions.

• The derivative of \( \ln x \) is \( \frac{1}{x} \).
• The integral of \( \frac{1}{x} \) is \( \ln |x| + C \), where \( C \) is the constant of integration.

In integration by parts, as seen in this exercise, we choose \( u = \ln x \) because its derivative simplifies our calculations. The natural logarithm plays a critical role in areas of calculus concerning exponential growth and decay, making it indispensable in both theoretical and applied contexts.