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Magazine Chapter 6: Problem 59
Area Find the area under the curve \(y=x \ln x\) and above the \(x\) -axis from \(x=1\) to \(x=2\)
Short Answer
Expert verified
The area is \(2 \ln 2 - 0.75\).
Step by step solution
01
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 \([1, 2]\).
02
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 \(x = 2\).
03
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: \[\frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx.\]
04
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.
05
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: \( (2 \ln 2 - 1) - (0 - 0.25) = 2 \ln 2 - 0.75.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \( \frac{x^2}{2} \).
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.
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.
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.
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.
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.
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
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.
- Breaking down complex problems,
- Choosing the appropriate techniques such as integration by parts, and
- Accurately evaluating the results.
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.
