Question

Integrating a Discontinuous Integrand Evaluate \(\int_{0}^{4} \frac{1}{\sqrt{4-x}} d x\), if possible. State whether the integral converges or diverges.

Short answer

The integral converges to 4.

Step-by-step solution

Understanding the Discontinuity

The integrand \( \frac{1}{\sqrt{4-x}} \) is discontinuous at \( x = 4 \) because as \( x \to 4 \), \( \sqrt{4-x} \to 0 \), making the denominator approach zero and the integrand approach infinity.

Set Up the Improper Integral

Since the integrand is discontinuous at the upper limit of integration \( x = 4 \), this is an improper integral. We will evaluate \( \lim_{b \to 4^-} \int_0^b \frac{1}{\sqrt{4-x}} \, dx \).

Find the Antiderivative

The antiderivative of \( \frac{1}{\sqrt{4-x}} \) is \(-2 \sqrt{4-x} \). To find this, note that if \( u = 4-x \), then \( du = -dx \). Thus, the antiderivative becomes \( \int \frac{-1}{\sqrt{u}} \, du = -2 \sqrt{u} + C \).

Substitute and Evaluate the Integral

Calculate \( \lim_{b \to 4^-} \left[ -2 \sqrt{4-x} \right]_0^b = \lim_{b \to 4^-} \left( -2 \sqrt{4-b} + 2 \sqrt{4} \right). \) Evaluate the expression at the bounds: \(-2 \sqrt{4-b}\) as \(b \to 4^- \), approaches 0, and \(2 \times 2 = 4\).

Solve the Limit

Since \( -2 \sqrt{4-b} \) approaches \( 0 \) as \( b \to 4^- \), the limit becomes \( 4 \), the expression simplifies to \( 4 - 0 = 4 \).

Conclusion on Convergence

Since the limit exists and is finite, the improper integral converges to \( 4 \).

Key concepts

Discontinuous Functions

Discontinuous functions are those that have breaks, holes, or points of undefined values within their domain. In calculus, identifying discontinuities is crucial because they affect whether and how we can integrate a function. Consider the function \( \frac{1}{\sqrt{4-x}} \). As \( x \) approaches 4 from the left, the denominator nears zero, causing the function to shoot upwards to infinity. Thus, the function becomes undefined and discontinuous at this point.
This discontinuity at \( x = 4 \) affects the evaluation of the integral from 0 to 4. Understanding the nature of discontinuities helps us determine whether an integral is improper and how to handle it. Discontinuities can occur at either or both limits of integration or even within the interval of integration.
Handling these kinds of problems often involves setting up a limit process to "tame" the infinity or undefinable point, enabling us to evaluate the integral accurately.

Antiderivatives

An antiderivative of a function is another function whose derivative returns the original function. Essentially, it is the reverse process of differentiation. When integrating, finding an antiderivative allows us to evaluate definite integrals, especially with complex functions.
For the function \( \frac{1}{\sqrt{4-x}} \), the antiderivative is \(-2 \sqrt{4-x} \). To determine this, a substitution method is used. If we let \( u = 4-x \), then the differential \( du = -dx \). Plugging this into the integral gives us \( \int \frac{-1}{\sqrt{u}} \, du = -2 \sqrt{u} + C \), where \( C \) is the constant of integration.
This antiderivative is essential for evaluating the definite integral and tackling the limit necessary for improper integrals.

Limits of Integration

When working with definite integrals, limits of integration define the interval over which the function is being integrated. Typically, these limits are straightforward, such as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) indicate the bounds. However, improper integrals involve at least one of the limits causing undefined behavior or approaching infinity.
In the given exercise, the upper limit of 4 makes \( \frac{1}{\sqrt{4-x}} \) improper, as it causes a discontinuity. Thus, we set up an integral from 0 to \( b \) and look at the limit as \( b \to 4^- \). This process of taking a limit helps redefine the problematic endpoint in a reliable way for evaluation.
Understanding limits of integration is crucial in setting up and solving both standard and improper integrals effectively.

Convergence and Divergence

Convergence and divergence are crucial in determining the behavior of improper integrals. An integral converges if its evaluation leads to a finite number. Conversely, it diverges if it leads to an infinite result. Hence, analysis using limits is essential for determining the outcome of improper integrals.
For \( \int_{0}^{4} \frac{1}{\sqrt{4-x}} \, dx \), evaluating the improper integral involves a limit as the variable approaches the point of discontinuity. As seen, \( \lim_{b \to 4^-} \left[ -2 \sqrt{4-x} \right]_0^b \). This manipulation leads to a finite result of 4, as \( -2 \sqrt{4-b} \) approaches zero while the constant remains at 4.
The finite result signifies that the integral converges. Complete understanding of whether an integral converges or diverges is essential, particularly with discontinuous functions, to ensure meaningful interpretations of area under a curve.