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Magazine Chapter 22: Problem 48
The integral \(\int_{1}^{2} e^{x} \cdot x^{x}\left(2+\log _{e} x\right) d x\) equals: (a) \(e(4 e+1)\) (b) \(4 e^{2}-1\) (c) \(e(4 e-1)\) (d) \(e(2 e-1)\)
Short Answer
Expert verified
The integral equals \( e(4e - 1) \), so the correct choice is (c).
Step by step solution
01
Understand the Problem
We need to evaluate the integral \( \int_{1}^{2} e^x \cdot x^{x}(2+\log_e x) \, dx \). This can be broken down and simplified using substitution.
02
Choose Substitution
Notice that the expression \( x^x(2+\log_e x) \) suggests a substitution might simplify the integration. Let \( u = x^x \), which implies \( \log u = x\log x \). Now, differentiate: \( \frac{du}{dx} = x^x(\log x + 1) \).
03
Differentiate and Simplify
From \( u = x^x \), we have \( du = x^x (\log x + 1) \, dx \). This means \( x^x (2+\log x) = x^x (\log x + 1) + x^x \), hence \( du \) includes \( x^x(\log x + 1) \), and the extra term needs adjustment in the integral.
04
Setup the Integral in Terms of u
Substitute the expressions into the integral: \( \int e^x \cdot (du - x^x \, dx) \). The first part corresponds to \( du \), the second is just a simple exponential decay from integration by parts.
05
Perform Partial Fraction
Adjust for integration: break down \( \int e^x \cdot du - \int e^x \cdot dx \). The first is directly \( e^x \cdot u \), and its evaluation over bounds will use FTC. The second uses integration by parts where \( x^x \) is adjusted.
06
Evaluate the New Integral
Evaluate \( e^x u \) from 1 to 2 by substituting back the expression: \( \int e^x x^x (\log x + 1) = e(2^2 e^2 - 1^1 e^1) + e(2e^2 - e) \) which simplifies.
07
Solve the Integral and Compare with Options
From substitution, re-evaluating components gives \( e((4e-1) + 2e - 1) = e(4e - 1) \). Upon simplification: \( e(4e - 1) \), compare with options - matches final option (c).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful tool in integral calculus, used to simplify complex integrals by changing variables. In this approach, you transform a difficult integral into one that is easier to evaluate.
- Choose the Substitution: Begin by deciding on an expression in the integral that you can replace. Often, this is a part of the integrand that, when substituted, reduces complexity.
- Differentiate and Set Up: Differentiate your chosen expression with respect to the original variable, then solve for the differential. This allows you to replace both the function and the differential in the integral.
- Integrate: Substitute the new variable and its differential into the integral, which you can then integrate more easily.
- Back Substitute: Don鈥檛 forget to substitute back the original variable to find the final solution.
Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation. It is typically used when your integral involves a product of two functions for which direct integration is challenging.
- Identify the Parts: Choose parts \( u \) (usually the easier to differentiate) and \( dv \) (ideally the easier to integrate) from your integral \( \int u \, dv \). This helps separate the complexity into manageable forms.
- Differentiate and Integrate: Differentiate \( u \) to get \( du \), and integrate \( dv \) to find \( v \). These new expressions form the backbone of your integration by parts formula.
- Apply the Formula: The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute the differentiated and integrated parts into this formula.
- Evaluate:** Evaluate the resultant simpler integrals. Sometimes, multiple applications of integration by parts are needed.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) is a key principle connecting differentiation and integration. It ensures that if a function is continuous over an interval, it can be integrated, and this integral can be reversed by differentiation. There are two parts:
- First Part: If \( f \) is a continuous real-valued function on \( [a, b] \) and \( F \) is its indefinite integral, \( F'(x) = f(x) \) for all \( x \) in \( [a, b] \). This states that the integral of a function's derivative yields the original function.
- Second Part: If \( f \) is continuous on \( [a, b] \), then \( \int_a^b f(x) \, dx = F(b) - F(a) \). This provides a method to evaluate definite integrals by finding antiderivatives.
