Chapter 14: Problem 37
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 \[ F(b) - F(a) \]
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 \( x \).
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:
- \[ \int u \, dv = uv - \int v \, du \]
- 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.