Problem 115 - University Calculus: Early Transcendentals

Question

If \(f\) is a continuous function, find the value of the integral $$I=\int_{0}^{a} \frac{f(x) d x}{f(x)+f(a-x)}$$ by making the substitution \(u=a-x\) and adding the resulting integral to \(I\)

Step-by-Step Solution

Answer

The value of the integral \(I\) is \(\frac{a}{2}\).

Step 1: Original Integral

Consider the given integral: \[I = \int_{0}^{a} \frac{f(x) \, dx}{f(x) + f(a-x)}\].Our goal is to evaluate this integral using substitution.

Step 2: Substitution

Perform the substitution \(u = a - x\), which implies \(du = -dx\). Rewrite the limits of integration: when \(x = 0\), \(u = a - 0 = a\), and when \(x = a\), \(u = a - a = 0\). The integral becomes:\[\int_{a}^{0} \frac{f(a-u)(-du)}{f(a-u) + f(u)}.\] Since reversing the limits changes the sign, we rewrite it as:\[\int_{0}^{a} \frac{f(a-u) \, du}{f(a-u) + f(u)}.\]

Step 3: Add the Integrals

We now have two integrals:\[I = \int_{0}^{a} \frac{f(x) \, dx}{f(x) + f(a-x)}\]and\[J = \int_{0}^{a} \frac{f(a-u) \, du}{f(a-u) + f(u)}\].Add these to solve for \(I + J\):\[I + J = \int_{0}^{a} \left( \frac{f(x)}{f(x) + f(a-x)} + \frac{f(a-x)}{f(a-x) + f(x)} \right) dx.\] Since the denominators are the same, they simplify to 1:\[ I + J = \int_{0}^{a} (1) \, dx.\]

Step 4: Evaluate the Integral

The integral of 1 over the interval \([0, a]\) is straightforward:\[I + J = \int_{0}^{a} (1) \, dx = [x]_{0}^{a} = a.\]Since the substitution technique reveals that \(I = J\), we have:\[2I = a.\]

Step 5: Solve for I

From \(2I = a\), solve for \(I\):\[I = \frac{a}{2}.\]

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify integrals. It involves substituting a part of the integral with a new variable, making the integral easier to evaluate. In this exercise, we used the substitution \(u = a - x\).

Here's why this substitution works beautifully:

• First, by substituting \(u = a - x\), we transform the variable of integration.
• This changes the differential from \(dx\) to \(-du\), and also flips the integration limits from \(0\) to \(a\), and \(a\) to \(0\).

Reversing limits of integration switches the sign of the integral, which is a crucial step in simplifying the expression. Through the substitution, the problem often reveals symmetries or identical terms that can be combined or made simpler, as it did in this particular example.

Continuous Function

A continuous function is one that has no breaks, jumps, or gaps in its domain where it is defined. In calculus, dealing with continuous functions is very important because it allows us to apply various integration techniques, such as the substitution method seen in this exercise.

Continuous functions have some helpful properties:

• They are predictable and smooth over their entire domain.
• You can compute definite integrals directly using lower and upper limits without worrying about points of discontinuity.

In the exercise, the continuity of the function \(f(x)\) ensures that our substitution and the subsequent integrals are both valid and straightforward. Because \(f(x)\) does not have any discontinuities, the calculations proceed smoothly, and the integrity of our solution is maintained.

Definite Integral

A definite integral refers to the evaluation of an integral with specific upper and lower limits, often representing the area under a curve between two points. In this exercise, the integral \(\int_{0}^{a} \frac{f(x) \, dx}{f(x) + f(a-x)}\) is a definite integral.

Here's what makes definite integrals special:

• They provide a numerical value, as opposed to indefinite integrals that result in a general form plus a constant.
• The evaluation is bounded by the upper and lower limits, which in our example are \(0\) and \(a\).

The application of definite integrals in our solution is essential, as adding the two modified integrals and simplifying back to a constant \(a\) demonstrates their utility in evaluating a problem to a precise answer, \(I = \frac{a}{2}\).

Integral Calculus

Integral calculus focuses on the concepts of integrals and their application, which involve finding areas under curves and solving differential equations among many other things. It is one of the two main branches of calculus, the other being differential calculus.

The importance of integral calculus includes:

• Its ability to evaluate quantities such as area, volume, and the accumulated quantity from a rate of change.
• Providing solutions to problems involving the inverse process of differentiation, allowing us to recover original functions from their derivatives.

In this exercise, integral calculus was used to solve an integral involving a continuous function \(f(x)\) and demonstrate how transformations and symmetries can simplify the computation. This highlights the ultimate goal of integral calculus: to provide tools and methods to solve for unknown quantities efficiently, revealing deeper mathematical principles such as how \(I\) reduced to \(\frac{a}{2}\).