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Chapter 4: Problem 17

Evaluate using integration by parts or substitution. Check by differentiating. $$\int(x-1) \ln x \, d x$$

Short Answer

Expert verified
\(\ln x \left( \frac{x^2}{2} - x \right) - \left( \frac{x^2}{4} - x \right) + C\).

Step by step solution

01

- Identify parts for integration by parts

Integration by parts formula is: \[\int u \, dv = uv - \int v \, du\]Select \(u\) and \(dv\). Let \(u = \ln x\) and \(dv = (x-1) \, dx\).
02

- Compute derivatives and integrals

Compute \(du\) and \(v\):\[du = \frac{d}{dx} (\ln x) \, dx = \frac{1}{x} \, dx,\]\[v = \int (x-1) \, dx = \int x \, dx - \int 1 \, dx = \frac{x^2}{2} - x.\]
03

- Apply integration by parts

Substitute \(u\), \(dv\), \(v\), and \(du\) into the integration by parts formula:\[\int (x-1) \, \ln x \, dx = \ln x \left( \frac{x^2}{2} - x \right) - \int \left( \frac{x^2}{2} - x \right) \frac{1}{x} \, dx.\]
04

- Simplify and integrate

Simplify the integral:\[= \ln x \left( \frac{x^2}{2} - x \right) - \int \left( \frac{x}{2} - 1 \right) \, dx.\]Integrate each term separately:\[\int \left( \frac{x}{2} - 1 \right) \, dx = \int \frac{x}{2} \, dx - \int 1 \, dx = \frac{x^2}{4} - x.\]
05

- Substitute back and combine terms

Combine all terms to get the final answer:\[\int (x-1) \, \ln x \, dx = \ln x \left( \frac{x^2}{2} - x \right) - \left( \frac{x^2}{4} - x \right) + C.\]
06

- Check by differentiating

Differentiate the answer to check if it matches the original integrand:\[d\left( \ln x \left( \frac{x^2}{2} - x \right) - \left( \frac{x^2}{4} - x \right) + C \right) / dx\]Should simplify back to \((x-1) \ln x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
A definite integral is an integral with specified upper and lower limits. It provides the net area under a curve between these two limits. For a function \( f(x) \), the definite integral from \( a \) to \( b \) is denoted as: \[\begin{equation*}\text{\textbackslash{}int}_{a}^{b} f(x) \, dx.\text{\textbackslash{}end{equation*}\]Unlike indefinite integrals, definite integrals result in a number, not a function. To evaluate a definite integral:
  • Find the antiderivative of \( f(x) \) and denote it as \( F(x) \).
  • Compute \( F(b) - F(a) \).
If dealing with integration by parts or substitution as in our original exercise, you first solve the indefinite integral and then apply the limits.
Differentiation
Differentiation is the process of finding the derivative of a function. The derivative represents how a function changes as its input changes. It's a fundamental concept in calculus. For example, the derivative of \( f(x) = x^2 \) is \( f'(x) = 2x \). The differentiation rules you'll frequently use include:
  • The Power Rule: \( \frac{d}{dx} x^n = nx^{n-1} \)
  • The Product Rule: \( (uv)' = u'v + uv' \)
  • The Chain Rule: \( \frac{d}{dx} f(g(x)) = f'(g(x))g'(x) \)
In our step-by-step solution, differentiation was used to check the correctness of the integral result. By differentiating the final expression, we should retrieve the original integrand. This confirms our integration was performed correctly.
Natural Logarithm
The natural logarithm, denoted as \( \ln x \), is a logarithm with base \( e \), where \( e \) approximates to 2.71828. It has several key properties:
  • \( \ln(1) = 0 \)
  • \( \ln(ab) = \ln(a) + \ln(b) \)
  • \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
  • \( \ln(a^b) = b \cdot \ln(a) \)
When performing integration by parts with natural logarithms, it is common to choose \( u = \ln x \) because its derivative, \( \frac{1}{x} \), simplifies the integral. Notice in our original solution, we let \( u = \ln x \) and then followed with the differentiation: \( du = \frac{1}{x} \, dx \). Coupled with other factors, this selection aids in systematically simplifying and solving the integral effectively.

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Most popular questions from this chapter

Industrial learning curve. A company is producing a new product. Due to the nature of the product, the time required to produce each unit decreases as workers become more \intamiliar with the production procedure. It is determined that the function for the learning process is $$T(x)=2+0.3\left(\frac{1}{x}\right)$$ where \(T(x)\) is the time, in hours, required to produce the \(x\) th unit. Use this information. Find the total time required for a new worker to produce units 1 through \(10 ;\) units 20 through 30

Evaluate. Assume \(u>0\) when ln u appears. $$\int \frac{t^{3} \ln \left(t^{4}+8\right)}{t^{4}+8} d t$$

Find the area under the graph of each function over the given interval. $$y=e^{x} ; \quad[-2,3]$$

Evaluate. Assume \(u>0\) when ln u appears. $$\int \frac{d x}{e^{x}+1}\left(\text { Hint: } \frac{1}{e^{x}+1}=\frac{e^{-x}}{1+e^{-x}}\right)$$

Evaluate. $$\int_{-2}^{2} \sqrt{4-x^{2}} d x$$
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