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Magazine Chapter 7: Problem 34
Evaluate the integrals in Exercises \(29-50.\) $$\int 2 e^{(2 x-1)} d x$$
Short Answer
Expert verified
The integral evaluates to \( e^{(2x-1)} + C \).
Step by step solution
01
Identify the Integral
The given integral is \( \int 2 e^{(2x-1)} \, dx \). This is an exponential integral where the integrand is of the form \( e^{ax+b} \), with \( a = 2 \) and \( b = -1 \).
02
Determine Substitution Function
Let \( u = 2x - 1 \). This substitution will simplify the integral.
03
Find Derivative of Substitution
Differentiate \( u = 2x - 1 \) with respect to \( x \), which gives \( \frac{du}{dx} = 2 \). That means \( du = 2 \, dx \).
04
Solve for dx in terms of du
From \( du = 2 \, dx \), solve for \( dx \): \( dx = \frac{1}{2} \ du \).
05
Substitute in the Integral
Substituting \( u \) and \( dx \) in the original integral: \( \int 2 e^{(2x-1)} \, dx = \int 2 e^u \cdot \frac{1}{2} \, du = \int e^u \, du \).
06
Evaluate the New Integral
The integral \( \int e^u \, du \) is simply \( e^u + C \), where \( C \) is the constant of integration.
07
Back Substitute u
Substitute back \( u = 2x - 1 \) into \( e^u + C \) to get \( e^{(2x-1)} + C \).
08
Final Answer
Thus, the solution to the integral \( \int 2 e^{(2x-1)} \, dx \) is \( e^{(2x-1)} + C \).
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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 crucial technique in calculus for solving integrals, especially when dealing with complex expressions. Essentially, this method involves transforming the original integral into a simpler form that is easier to evaluate. Here鈥檚 a step-by-step breakdown:
- First, identify a part of the integrand that can be substituted with a single variable to simplify the integral. In our case, it's the expression inside the exponent: \( 2x - 1 \).
- This expression is replaced with a new variable \( u = 2x - 1 \), which serves to simplify the expression \( e^{2x-1} \) into \( e^u \).
- Then, find the derivative of \( u \) with respect to \( x \) to determine the necessary adjustment for \( dx \), here \( \frac{du}{dx} = 2 \).
- Solve this derivative equation for \( dx \), yielding \( dx = \frac{1}{2} \, du \), and use this in your integral.
Indefinite Integrals
Indefinite integrals represent the process of finding the antiderivatives of a given function. Unlike definite integrals, which produce numerical values, indefinite integrals result in a family of functions. Here are some key points to understand:
- The result of an indefinite integral includes a "constant of integration," denoted as \( C \). This constant accounts for all the possible antiderivatives that differ by a constant.
- The notation \( \int f(x) \, dx \) represents the indefinite integral of \( f(x) \) with respect to \( x \).
- In our example, once the substitution method simplifies the integral to \( \int e^u \, du \), solving it involves recognizing that the antiderivative of \( e^u \) is \( e^u + C \).
- After finding the antiderivative, it is crucial to back-substitute the original variable to express the result in terms of the initial variable \( x \).
Integration Techniques
Mastering various integration techniques is vital for tackling a wide array of problems in mathematics. Each technique is useful under different scenarios, and familiarity with them can drastically simplify problem-solving:
- Substitution: As seen in our example, substitution is very useful when dealing with composite functions, particularly when an integral seems too complex to tackle directly.
- Integration by Parts: Useful when the integrand is a product of two different types of functions, like polynomials and exponentials.
- Partial Fractions: Effective for breaking down rational functions into simpler parts that can be easily integrated.
- Trigonometric Integrals and Substitutions: Especially handy for integrals involving trigonometric functions or when converting rational expressions to easier trigonometric equivalents.
