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Magazine Chapter 5: Problem 90
Suppose that \(f\) is continuous on \([a, b] .\) let \(\delta=(b-a) / 2\) and \(\mu=(a+b) / 2\). Use a substitution to show that $$ \int_{a}^{b} f(u) d u=\delta \int_{-1}^{1} f(\delta x+\mu) d x $$
Short Answer
Expert verified
Use the substitution \( u = \delta x + \mu \) to transform the integral bounds and differential, resulting in the desired integral form.
Step by step solution
01
Understanding the Problem
We need to show that the integral of a function \( f(u) \) from \( a \) to \( b \) can be transformed into an integral from \( -1 \) to \( 1 \) using a substitution method.
02
Define Required Substitution Variables
Let us define \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \), which define the transformation we need for changing the interval from \( [a, b] \) to \( [-1, 1] \).
03
Perform the Substitution
First, substitute \( u = \delta x + \mu \). Then differentiate with respect to \( x \): \( du = \delta dx \).
04
Change Integration Limits
Since we have \( u = \delta x + \mu \), when \( u = a \), \( x = -1 \) (because \( a = \mu - \delta \)). When \( u = b \), \( x = 1 \) (since \( b = \mu + \delta \)).
05
Substitute in Original Integral
Substitute \( u = \delta x + \mu \) and \( du = \delta dx \) into the original integral: \[ \int_{a}^{b} f(u) \, du = \int_{-1}^{1} f(\delta x + \mu) \delta \, dx \]
06
Factor and Simplify
Factor out the constant \( \delta \) from the integral:\[ \int_{a}^{b} f(u) \, du = \delta \int_{-1}^{1} f(\delta x + \mu) \, dx \] This shows the desired transformation.
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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 technique for simplifying integrals. It works by replacing a variable in the integral with another variable, usually to transform the integration into a simpler or more recognizable form. In our specific problem, we're substituting the variable \( u \) with \( \delta x + \mu \).
This substitution simplifies the process by aligning the limits of integration with a more symmetric interval \([-1, 1]\). It essentially "re-labels" the input values, which can make solving the integral more convenient.
Here's how it applies in our example:
This substitution simplifies the process by aligning the limits of integration with a more symmetric interval \([-1, 1]\). It essentially "re-labels" the input values, which can make solving the integral more convenient.
Here's how it applies in our example:
- We started with the integral \( \int_{a}^{b} f(u) \, du \).
- By substituting \( u = \delta x + \mu \) (where \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \)), we adjust the formula accordingly.
- This leads to the new integral representation \( \int_{-1}^{1} f(\delta x + \mu) \, \delta \, dx \), demonstrating the utility and elegance of the substitution method.
Change of Interval
The change of interval is a specific application of the substitution method. It modifies the range of integration to transform an integral into a more manageable form. In the given problem, the original interval \([a, b]\) is changed to \([-1, 1]\).
This is achieved by recognizing that any shifted interval \([a, b]\) can be transformed using \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \). Here's how it works:
This is achieved by recognizing that any shifted interval \([a, b]\) can be transformed using \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \). Here's how it works:
- As the substitution \( u = \delta x + \mu \) is applied, the limits of integration move.
- With \( u = a \) leading to \( x = -1 \) and \( u = b \) giving \( x = 1 \), the integral interval is now symmetric around the origin.
- This symmetric interval helps in simplifying certain types of functions, especially those like polynomials or sine and cosine functions.
Continuous Function
A continuous function is one that has no breaks, jumps, or gaps in its domain. For integration, especially definite integrals, continuity is crucial because it ensures the integral is well-defined over the interval. In our problem, the function \( f(u) \) is continuous on \([a, b]\).
Continuity allows us to perform operations like substitution and change of intervals without worrying about discontinuities that could disrupt these transformations.
In practical terms:
Continuity allows us to perform operations like substitution and change of intervals without worrying about discontinuities that could disrupt these transformations.
In practical terms:
- A continuous function behaves predictably, allowing us to apply limits and transformations smoothly.
- If \( f \) were not continuous, the integral might encounter undefined sections or infinities that would make calculation difficult or impossible.
Definite Integrals
A definite integral calculates the "net area" under a curve within specified bounds on the x-axis. For example, \( \int_{a}^{b} f(u) \, du \) evaluates the area from \( a \) to \( b \).
Definite integrals are a central concept in calculus and focus on finding total accumulation or change over an interval.
In our substitution example:
Definite integrals are a central concept in calculus and focus on finding total accumulation or change over an interval.
In our substitution example:
- The transformation from \( \int_{a}^{b} f(u) \, du \) to \( \delta \int_{-1}^{1} f(\delta x + \mu) \, dx \) is a re-expression but still computes this net area.
- Despite different integration limits, both expressions represent the same quantity 鈥� emphasizing that the value remains unchanged by substitution, only the integration process does.
