Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove (by a substitution) that $$ \int_{a}^{b} f(-x) d x=\int_{-b}^{-a} f(x) d x $$

Step-by-Step Solution

Answer

By substituting \( u = -x \), we showed \( \int_{a}^{b} f(-x) \, dx = \int_{-b}^{-a} f(x) \, dx \).

Step 1: Understand the Problem

We need to prove that \( \int_{a}^{b} f(-x) \, dx = \int_{-b}^{-a} f(x) \, dx \). The objective is to find a suitable substitution that will allow us to transform one integral into the other.

Step 2: Choose a Substitution

Select the substitution \( u = -x \). This substitution simplifies the expression because it directly replaces \( -x \) with \( u \).

Step 3: Differentiate the Substitution

Differentiate the substitution \( u = -x \) with respect to \( x \) to find \( du \):\[ du = -dx \]

Step 4: Adjust the Integration Limits

Change the limits of integration according to the substitution \( u = -x \):- When \( x = a \), \( u = -a \).- When \( x = b \), \( u = -b \).

Step 5: Substitute into the Integral

Substituting \( u = -x \) and \( du = -dx \) into the original integral \( \int_{a}^{b} f(-x) \, dx \):\[ \int_{a}^{b} f(-x) \, dx = \int_{-a}^{-b} f(u) (-du) \]

Step 6: Simplify the Integral

Recognize that the \(-\) sign outside changes the limits:\[ \int_{-a}^{-b} f(u) \, (-du) = \int_{-b}^{-a} f(u) \, du \]

Step 7: Conclusion

Therefore, by substitution, we have shown that:\( \int_{a}^{b} f(-x) \, dx = \int_{-b}^{-a} f(x) \, dx \).

Key Concepts

definite integrals

A definite integral represents the signed area under a curve between two specified points on the x-axis. In mathematical terms, if you have a function \( f(x) \) and you want to calculate the area between \( x = a \) and \( x = b \), you evaluate the definite integral \( \int_{a}^{b} f(x) \, dx \). This process provides you with the accumulated value from \( a \) to \( b \) of whatever \( f(x) \) represents.

• Definite integrals are crucial in finding areas, solving physics problems, and more.
• One important aspect of definite integrals is the limits of integration, which are \( a \) and \( b \) in \( \int_{a}^{b} \).

A distinctive feature of definite integrals is the importance of limits. These bounds convert the integral into a precise numerical value, distinguishing it from an indefinite integral, typically used for a more general solution. Adjusting these limits, as seen in the original exercise, is a fundamental skill when proving equalities of different integral expressions.

change of variables

The change of variables, also known as substitution, is a popular technique to simplify integrals, especially when dealing with complicated functions. This method involves substituting part of an integral with a new variable, transforming the integral into a potentially simpler form.

• To make a change of variables, identify a substitution \( u = g(x) \), replacing the original variable.
• Differentiate the substitution to find \( du \), which often involves rewiring \( dx \) in terms of \( du \).

For example, in the original exercise, by setting \( u = -x \), the integral of \( f(-x) \) turned into \( \int f(u) \, (-du) \), demonstrating the elegance of this technique in manipulating limits and simplifying the integral. Adjusting the limits is crucial: the bounds need to reflect the changes made by the substitution. This ensures consistency, allowing a seamless transition from one form of the integral to another.

integration techniques

Diverse integration techniques are available to solve integrals that don't easily succumb to standard methods. Integration by substitution is one such method and is often the first line of attack for rewriting and solving integrals.

• A popular integration technique, substitution, is essentially reverse chain rule application.
• Other techniques include integration by parts, partial fraction decomposition, and trigonometric substitution.

The choice of technique often depends on the form of the function within the integral. If you can recognize a function as a derivative of another (possibly up to a constant), substitution might simplify it significantly. Each technique has its own scenarios where it shines:

• For instance, integration by parts is useful for the product of functions such as polynomials and exponentials.
• Trigonometric substitution is typically employed when dealing with integrands involving square roots.

Understanding when and how to use these techniques enhances one's ability to tackle a wide range of problem types in calculus.