Problem 59 - Applied Calculus

Question

Area Find the area under the curve \(y=x \ln x\) and above the \(x\) -axis from \(x=1\) to \(x=2\)

Step-by-Step Solution

Answer

The area is \(2 \ln 2 - 0.75\).

Step 1: Identify Integral Limits and Function

The exercise asks for the area under the curve from \(x = 1\) to \(x = 2\). The function given is \(y = x \ln x\). So, we need to compute the definite integral of \(y = x \ln x\) over \([1, 2]\).

Step 2: Set Up the Definite Integral

The definite integral we need to evaluate is \(\int_{1}^{2} x \ln x \, dx\). This integral will provide the area under the curve \(y = x \ln x\) from \(x = 1\) to \(x = 2\).

Step 3: Integration by Parts

To solve \(\int x \ln x \, dx\), apply integration by parts. Select \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\). Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\), the integral becomes: \[\frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx.\]

Step 4: Simplify and Integrate

The integral from Step 3 simplifies to \(\int \frac{x}{2} \, dx\). This results in \(\frac{x^2}{4} + C\) upon integrating. Substitute back into the integration by parts formula: \[\frac{x^2}{2} \ln x - \left(\frac{x^2}{4}\right) \] and evaluate from 1 to 2.

Step 5: Evaluate Definite Integral

Calculate the integral from Step 4 at the bounds 2 and 1: \[\left(\frac{2^2}{2} \ln 2 - \frac{2^2}{4}\right) - \left(\frac{1^2}{2} \ln 1 - \frac{1^2}{4}\right).\] Simplifying the expression, we get: \( (2 \ln 2 - 1) - (0 - 0.25) = 2 \ln 2 - 0.75.\)

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to solve integrals that are products of functions, such as the integral of \(x \ln x\). This method is rooted in the product rule of differentiation. It allows us to break down a complex integral into simpler parts:

The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\). To apply it successfully:

• Choose \( u \) as the component that becomes simpler when differentiated. In our exercise, \( u = \ln x \) because differentiating \(\ln x\) yields \(\frac{1}{x}\), which is simpler.
• The remaining part, \( dv \), in our case \( x \, dx \), is then integrated to find \( v \), which is \( \frac{x^2}{2} \).

Integrating by parts can transform a challenging integral into two easier parts, helping us solve otherwise difficult calculus problems.

Area Under a Curve

Finding the area under a curve is a fundamental use of definite integrals in calculus. The area under a graph of a function between two limits gives the total accumulation, often representing physical quantities like distance or mass. In our specific exercise, we calculated the area under the curve \(y = x \ln x\) from \(x = 1\) to \(x = 2\).

By using a definite integral, \(\int_{1}^{2} x \ln x \, dx\), the solution captures the exact area. This is visualized as the region between the curve and the x-axis, from the starting point to the stopping point of our interval. Definite integrals provide a precise way to find such areas within specified bounds.

Natural Logarithm

The natural logarithm, denoted by \(\ln x\), is a logarithm with the base of Euler's number \(e\), roughly equal to 2.718. It is a crucial mathematical function used often in calculus because of its well-known differentiation and integration properties.

For the function \(x \ln x\), the natural logarithm contributes significantly to the behavior of the graph, influencing both how it curves and where it intercepts the axes.

• When differentiating \(\ln x\), you obtain \(\frac{1}{x}\).
• When integrating expressions involving \(\ln x\), techniques like integration by parts become essential.

Understanding how the natural logarithm interacts with other functions aids in finding solutions to complex calculus problems.

Applied Calculus

Applied calculus focuses on using mathematical concepts and techniques to solve real-world problems. This field involves taking calculus theories and putting them into practice in areas like physics, engineering, economics, and beyond.

In the context of our exercise, we use calculus to determine the area under the curve. This could translate into practical applications such as computing efficiency, assessing material stress, or even economic forecasting.

• Being able to calculate integrals means predicting outcomes based on graphical models.
• Understanding these principles is invaluable in fields where you measure change or accumulation.

Applied calculus helps bridge theoretical math with everyday issues, offering solutions to practical challenges.

Calculus Problem Solving

Learning to solve calculus problems efficiently requires practice and understanding of various techniques, like the one used in our problem with integrals. Problem-solving in calculus often involves

• Breaking down complex problems,
• Choosing the appropriate techniques such as integration by parts, and
• Accurately evaluating the results.

These methods help in simplifying problems and finding solutions that are not immediately obvious otherwise.

In our problem, starting with the integral limits and function choice, then moving through the integration by parts, required a structured approach. Consistent practice with these techniques fortifies your calculus skills, enabling you to tackle a wide array of mathematical challenges effortlessly.