Problem 309 - Calculus Volume 1

Question

In the following exercises, use a change of variables to show that each definite integral is equal to zero. \(\int_{0}^{1} \frac{1-2 t}{\left(1+\left(t-\frac{1}{2}\right)^{2}\right)} d t\)

Step-by-Step Solution

Answer

The integral is zero due to symmetry over the interval [0, 1].

Step 1: Identify the Integral

We are given the integral \( \int_{0}^{1} \frac{1-2t}{1+(t-\frac{1}{2})^2} \, dt \). The goal is to find a substitution that simplifies the integration process or makes it evident that the integral equals zero.

Step 2: Choose an Appropriate Substitution

Let's use the substitution \( u = 1 - t \). This substitution is chosen because it involves a simple linear transformation and allows us to exploit symmetry in the interval.

Step 3: Transform the Limits of Integration

With the substitution \( u = 1 - t \), when \( t = 0 \), \( u = 1 \), and when \( t = 1 \), \( u = 0 \). This means the integral becomes \( \int_{1}^{0} \frac{1-2(1-u)}{1+((1-u)-\frac{1}{2})^2} \, (-du) \).

Step 4: Simplify the Expression

The integral \( \int_{1}^{0} \) can be rewritten as \(-\int_{0}^{1} \). So, we have: \( \int_{0}^{1} \frac{1-2 + 2u}{1+(\frac{1}{2} - u)^2} \, du \).

Step 5: Simplify the Integrand

The integrand becomes \( \frac{2u - 1}{1+(\frac{1}{2} - u)^2} \). Notice, this is the negative of the original integrand when evaluated over a symmetric interval.

Step 6: Utilize Symmetry to Conclude

The original integrand and the transformed integrand (\( \frac{2t - 1}{1+(t-\frac{1}{2})^2} \) and \( \frac{1 - 2u}{1+((1-u)-\frac{1}{2})^2} \)) are negatives of each other over the interval \([0, 1]\). This symmetry implies that the areas under the curve from 0 to 1 cancel each other out, thus the definite integral equals zero.

Key Concepts

Change of Variables

When dealing with complex integrals, using a change of variables can substantially simplify the process. This technique involves altering the variable of integration to transform the integral into a more manageable form.
It can reveal symmetries or even parts that cancel each other out.

• Choose a substitution that reflects a potential symmetry in the integral, such as reversing or shifting the variable.
• Apply this new substitution throughout the integral. Remember to also change the differential and update the limits according to the new variable.
• Evaluate the simplified integral. Often, integrating over symmetric bounds helps in identifying zero values or other simplifications in definite integrals.

In the example problem, substituting \(u = 1-t\) helped in recognizing the symmetry between \(t\) and \(1-t\) over the interval \([0, 1]\), ultimately showing that the integral equals zero due to canceling contributions.

Definite Integral

A definite integral calculates the net area under a curve for a given interval. Unlike an indefinite integral, which results in a general antiderivative, a definite integral provides a specific numeric value. This is crucial when determining the total change or net effect over a certain range.

• Essentially, definite integrals measure the accumulation of quantities, such as area under the curve, across an interval \([a, b]\).
• They can be interpreted as the limit of a sum of rectangle areas, which becomes exact as the rectangles become infinitesimally thin.
• The definite integral notation, \(\int_{a}^{b} f(x) \, dx\), comprises the integrand \(f(x)\), the limits of integration \(a\) and \(b\), and the variable of integration \(dx\).

By evaluating the integral \(\int_{0}^{1} \frac{1-2t}{1+(t-\frac{1}{2})^2} \, dt\), the limits indicate calculating the area between \(t = 0\) and \(t = 1\). Using a variable change reveals symmetries or cancellations that can pinpoint if this area potentially sums to zero.

Symmetry in Integration

Symmetry in integration can significantly simplify solving definite integrals. If an integrand exhibits symmetry with respect to its interval of integration, it can indicate simplifications such as equal and opposite areas.

• Symmetries may include even or odd functions, symmetry around the y-axis, or negation transformations over an interval.
• Recognizing symmetry allows for straightforward conclusions, such as a zero result when equal area contributions cancel each other out.
• Checking symmetry often involves substituting \(x\) with \(-x\) or another expression to see how the function behaves over its domain.

In the given exercise, the integrands \(\frac{1-2t}{1+(t-\frac{1}{2})^2}\) and its transformed counterpart over \([0, 1]\) behave as negatives. This symmetry indicates that as one part of the interval contributes positively to the area, the other contributes equally negatively, resulting in an integral that sums to zero.