Problem 55 - Calculus: Early Transcendental Functions

Question

As in exercise \(54,\) evaluate \(\int_{0}^{2} \frac{f(x+4)}{f(x+4)+f(6-x)} d x\) for any positive, continuous function \(f\) on [0,2]

Step-by-Step Solution

Answer

The result of the integral \(\int_{0}^{2} \frac{f(x+4)}{f(x+4)+f(6-x)} d x\) is 1 for any positive, continuous function \(f\).

Step 1: Substitution 1

Make the change of variable \(u = x + 4\) in the definite integral. This will change the limits of integration from \([0, 2]\) to \([4, 6]\). The integral then becomes \(\int_{4}^{6} \frac{f(u)}{f(u) + f(6 - (u - 4))} du\) which simplifies to \(\int_{4}^{6} \frac{f(u)}{f(u) + f(2 + u)} du\).

Step 2: Substitution 2

Make another change of variable using \(v = 6 - u\). The limits change from \([4, 6]\) to \([0, 2]\) but in reverse order as \(v\) decreases when \(u\) increases. So, the integral becomes \(\int_{2}^{0} \frac{f(6 - v)}{f(6 - v) + f(2 + (6 - v))} (-dv)\). Simplify this to get \(\int_{0}^{2} \frac{f(2 + v)}{f(2 + v) + f(6 - v)} dv\) after changing the limits of integral.

Step 3: Adding Integrals

Add the integrals obtained in step 1 and 2. The result is \(\int_{0}^{2} \frac{f(u)}{f(u) + f(2 + u)} du + \int_{0}^{2} \frac{f(2 + v)}{f(2 + v) + f(6 - v)} dv\). Notice here \(u\) and \(v\) are dummy variables for integration, so we can use \(x\) instead. This gives, \(2 \int_{0}^{2} \frac{f(x)}{f(x) + f(6 - x)} dx\).

Step 4: Simplify and Find Final Answer

Isolate the required integral in terms of other quantities from the result in step 3. Divide the equation by 2 to get, \(\int_{0}^{2} \frac{f(x)}{f(x) + f(6 - x)} dx = \frac{1}{2} \int_{0}^{2} dx\). This integral can be evaluated directly to obtain 1. Therefore, the solution of the original problem is 1.

Key Concepts

Integration Techniques

Integration can seem daunting at first, but it's critical to calculus problem-solving. Definite integrals, specifically, involve calculating the area under a curve within set boundaries.
A solid approach is necessary to handle these integrals, and one common method is substitution.

• Substitution: This technique involves changing the variable of integration. It helps simplify the integral by rewriting it in a more manageable form.
• Integration by Parts: Although not used here, it's another powerful technique where the product rule of differentiation is reversed.

In our specific exercise, substitution was the star. By using clever substitutions, the problem was reduced to its simplest form, making it easier to integrate. This shows just how versatile and helpful integration techniques can be when faced with complex calculus problems.

Change of Variables

The change of variables is an ingenious strategy used to simplify integrals.
It's all about choosing the right substitution to make the integral more straightforward or recognizable. In math terms, it's transforming the variable in an integral to a new variable.

• For example, in our problem, substituting \(u = x + 4\) modified the interval and function dramatically but kept the integral's value unchanged.
• Another substitution \(v = 6 - u\) continued the simplification.

These transformations helped manage the complexity and were pivotal in solving the integral neatly. Each substitution moves us closer to our solution, maintaining the equality while making calculations easier.

Calculus Problem Solving

Solving a calculus problem is like piecing together a puzzle.
It involves breaking down complex processes into manageable steps. Each step in the original solution brought us closer to finding the integral's value.

• Understand the Problem: Before diving into calculations, recognize the problem type and the applicable math strategies.
• Step-by-Step Solutions: Integrals can often be solved by splitting difficult expressions and progressively simplifying them. That was evident as the problem was systematically solved using change of variables.
• Verifying Results: Always check your work to see if the solution logically follows from the steps and meets any constraints given at the start of the problem.

In our exercise, the solution demonstrated that by carefully applying transformations and checking each step, complex calculus problems could be unraveled efficiently. This strategic approach is essential across all areas of calculus problem-solving.