/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 Evaluate the integrals using int... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals using integration by parts where possible. $$\int_{1}^{2} x \ln (2 x) d x$$

Short Answer

Expert verified
The short answer is: \(\int_{1}^{2} x \ln (2 x) d x = 4\ln(4) - \frac{1}{2}\ln(2) -\frac{3}{2}\)

Step by step solution

01

Choose u and dv

Let u(x) = ln(2x) and dv(x) = x dx.
02

Differentiate u and integrate dv

To differentiate u, we have: \(du=\frac{d}{dx}(ln(2x))\,dx\) Applying chain rule, we get: \(du=\frac{1}{x} \cdot 2 \, dx=\frac{2}{x}\,dx \) Next, integrate dv: \(v=\int x\,dx\) \(v=\frac{1}{2}x^2\)
03

Apply integration by parts formula

Now, use the integration by parts formula: \(\int_{1}^{2} x \ln (2 x) d x = \int_{1}^{2} u\, dv =UV-\int_{1}^{2} v\,du\) Substitute the values of u, du, and v from Steps 1 and 2: \(\int_{1}^{2} x \ln (2 x) d x =(\frac{1}{2}x^2\ln(2x))|_{1}^{2} - \int_{1}^{2} \frac{1}{2}x^2 \frac {2}{x} dx\)
04

Simplify the expression and find the integral

Simplify the expression: \(\int_{1}^{2} x \ln (2 x) d x =(\frac{1}{2}x^2\ln(2x))|_{1}^{2} - \int_{1}^{2} x dx\) Now, compute the definite integral of the simplified expression: \((\frac{1}{2}x^2\ln(2x))|_{1}^{2} = \frac{1}{2}(2^2\ln(4) - 1^2\ln(2))\) Evaluate the remaining integral: \(- \int_{1}^{2} x\, dx = - \frac{1}{2}x^2|_{1}^{2} = -\frac{1}{2}(2^2-1^2)= -\frac{3}{2}\)
05

Combine both parts and give the final answer

Combine the results: \(\int_{1}^{2} x \ln (2 x) d x = (\frac{1}{2}x^2\ln(2x))|_{1}^{2} - \int_{1}^{2} x dx =\frac{1}{2}(2^2\ln(4) - 1^2\ln(2)) - \frac{3}{2}\) So, the answer is: \(\int_{1}^{2} x \ln (2 x) d x = 4\ln(4) - \frac{1}{2}\ln(2) -\frac{3}{2}\)

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Key Concepts

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

Definite Integrals
Definite integrals are a fundamental concept in calculus that help us work out the total "accumulated" value of a function over a closed interval. This means you consider the area under the curve of the function from one point to another. In simpler terms, if you were tracking distance on a trip, a definite integral could tell you the total distance traveled between two times. To compute a definite integral, you choose two boundaries or limits, which we can denote as "a" and "b". You then integrate the function over this interval. The result gives a specific number, unlike indefinite integrals which leave you with a family of functions plus a constant C.
  • A definite integral can be represented as: \[ \int_{a}^{b} f(x) \, dx \]
  • The solution to this is essentially finding \[ F(b) - F(a) \]
For the given interval from 1 to 2, the problem involves determining the cumulative effect of the function \( x \ln(2x) \), capturing how it behaves within this range.
Chain Rule
The chain rule is a crucial differentiation technique, especially when dealing with composite functions. Essentially, it allows you to find the derivative of a function that is made by combining two or more simpler functions. In practice, think about cracking open layers of a mathematical onion; each layer being a separate function that needs to be differentiated. The chain rule is your guide to peel away these layers smoothly. Here’s how it generally works:
  • If you have a composite function, say \( f(g(x)) \), the chain rule states: \[ \frac{dy}{dx} = \frac{dy}{dg} \times \frac{dg}{dx} \]
  • This means you take the derivative of the outer function with respect to the inner function, then multiply it by the derivative of the inner function with respect to \( x \).
In our problem, using ln(2x) involved applying the chain rule because we were dealing with the natural logarithm of a scaled variable (2x), and thus ensured the proper differentiation was performed for subsequent calculations.
Integration Techniques
When faced with an integral, choosing the right technique can simplify the process significantly. Integration by parts is a powerful technique, particularly useful when you're working with products of functions like polynomials and logarithms or exponentials.This method stems from the product rule for differentiation and is formally expressed as:
  • \[ \int u \, dv = uv - \int v \, du \]
Here's a simple breakdown of how to select and apply this method:
  • Identify parts of the integrand to label as \( u \) and \( dv \).
  • Differentiate \( u \) to find \( du \).
  • Integrate \( dv \) to find \( v \).
  • Plug these into the integration by parts formula.
This technique is especially handy for definite integrals because you can directly solve and find the limits without having to deal with a constant of integration. In solving \( \int_{1}^{2} x \ln(2x) \, dx \), integration by parts allowed us to handle the logarithmic function efficiently alongside the variable \( x \).

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Most popular questions from this chapter

Decide whether or not the given integral converges. If the integral converges, compute its value. $$\int_{0}^{2} \frac{1}{x^{2}} d x$$

Market Saturation You have just introduced a new model of DVD player. You predict that the market will saturate at \(2,000,000\) DVD players and that your total sales will be governed by the equation $$ \frac{d S}{d t}=\frac{1}{4} S(2-S) $$ where \(S\) is the total sales in millions of DVD players and \(t\) is measured in months. If you give away 1,000 DVD players when you first introduce them, what will \(S\) be? Sketch the graph of \(S\) as a function of \(t .\) About how long will it take to saturate the market? (See Exercise 39.)

Epidemics A certain epidemic of influenza is predicted to follow the function defined by $$ \frac{d A}{d t}=\frac{1}{10} A(20-A) $$ where \(A\) is the number of people infected in millions and \(t\) is the number of months after the epidemic starts. If 20,000 cases are reported initially, find \(A(t)\) and sketch its graph. When is \(A\) growing fastest? How many people will eventually be affected? (See Exercise 39.)

Which of the following integrals are improper, and why? (Do not evaluate any of them.) a. \(\int_{-1}^{1} \frac{|x|}{x} d x \quad\) b. \(\int_{-1}^{1} x^{-1 / 3} d x \quad\) c. \(\int_{0}^{2} \frac{x-2}{x^{2}-4 x+4} d x\)

Sales Your monthly sales of Green Tea Ice Cream are falling at an instantaneous rate of \(5 \%\) per month. If you currently sell 1,000 quarts per month, find the differential equation describing your change in sales, and then solve it to predict your monthly sales. HINT [See Example 3.]

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