Zero net area Consider the function f(x) = (1 − x)/x
a. Are there numbers 0 < a < 1 such that ∫₁₋ₐ¹⁺ᵃ f(x) dx = 0?

Verified step by step guidance

1. First, understand the problem: we want to find if there exists a number \( a \) with \( 0 < a < 1 \) such that the definite integral \( \int_{1 - a}^{1 + a} f(x) \, dx = 0 \), where \( f(x) = \frac{1 - x}{x} \).
2. Rewrite the function \( f(x) \) to simplify integration: \( f(x) = \frac{1}{x} - 1 \). This allows us to split the integral into two simpler integrals.
3. Set up the integral as \( \int_{1 - a}^{1 + a} \left( \frac{1}{x} - 1 \right) dx = \int_{1 - a}^{1 + a} \frac{1}{x} \, dx - \int_{1 - a}^{1 + a} 1 \, dx \).
4. Evaluate each integral separately: \( \int \frac{1}{x} \, dx = \ln|x| \) and \( \int 1 \, dx = x \). Substitute the limits \( 1 - a \) and \( 1 + a \) into each result.
5. Combine the evaluated integrals to form the expression \( \left[ \ln|x| \right]_{1 - a}^{1 + a} - \left[ x \right]_{1 - a}^{1 + a} \). Set this equal to zero and solve the resulting equation for \( a \) in the interval \( (0,1) \).

Key Concepts

Definite Integral and Net Area

The definite integral of a function over an interval represents the net area between the function's graph and the x-axis. Positive areas above the axis add to the integral, while areas below subtract. Understanding how these areas combine is essential to determine when the integral equals zero.

Behavior of Rational Functions Near Singularities

The function f(x) = (1 - x)/x has a singularity at x = 0, where it is undefined. Analyzing the function's behavior near this point and within the interval around 1 is crucial to evaluate the integral and understand how the function's sign changes affect the net area.

Symmetry and Interval Selection in Integration

Choosing symmetric intervals around a point (here, around x=1) can lead to cancellation of positive and negative areas in the integral. Recognizing how the function behaves on [1 - a, 1 + a] helps determine if there exists an 'a' such that the integral sums to zero.