Problem 13 - Calculus for Biology and Medicine

Question

All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{-2}^{2} \frac{d x}{\sqrt{4-x^{2}}} $$

Step-by-Step Solution

Answer

The improper integral evaluates to \( \pi \).

Step 1: Identify Improper Integral

The integral \( \int_{-2}^{2} \frac{d x}{\sqrt{4-x^{2}}} \) is improper because the integrand \( \frac{1}{\sqrt{4-x^2}} \) becomes undefined at the endpoints \( x = 2 \) and \( x = -2 \), where the denominator becomes zero. This results in division by zero, indicating an improper integral.

Step 2: Express Integral as Limit

To handle the improper integral, we express it as a limit: \[ \lim_{a \to 2^-} \int_{-2}^{a} \frac{1}{\sqrt{4-x^2}} \, dx + \lim_{b \to -2^+} \int_{b}^{2} \frac{1}{\sqrt{4-x^2}} \, dx \]. This separates the integral into two parts and approaches each endpoint separately, considering their limit as the point is approached.

Step 3: Evaluate Indefinite Integral

Recognize that the integrand resembles the derivative of the arcsine function. The indefinite integral is \( \int \frac{1}{\sqrt{4-x^2}} \, dx = \arcsin\left(\frac{x}{2}\right) + C \). Here, \( \arcsin \) is utilized because its derivative is the reciprocal of \( \sqrt{4-x^2} \), similar to the integrand.

Step 4: Evaluate Definite Integrals

We now evaluate the definite integrals: \[ \lim_{a \to 2^-} \left[ \arcsin\left(\frac{x}{2}\right) \right]_{-2}^{a} + \lim_{b \to -2^+} \left[ \arcsin\left(\frac{x}{2}\right) \right]_{b}^{2} \]. This means calculating \( \arcsin\left(\frac{a}{2}\right) - \arcsin(-1) \) and \( \arcsin(1) - \arcsin\left(\frac{b}{2}\right) \).

Step 5: Apply Limits

As \( a \to 2^- \), \( \arcsin\left(\frac{a}{2}\right) \to \frac{\pi}{2} \) and \( \arcsin(-1) = -\frac{\pi}{2} \). As \( b \to -2^+ \), \( \arcsin(1) = \frac{\pi}{2} \) and \( \arcsin\left(\frac{b}{2}\right) \to -\frac{\pi}{2} \).

Step 6: Combine Results

Combine the evaluated limits: \( \left( \frac{\pi}{2} + \frac{\pi}{2} \right) + \left( \frac{\pi}{2} + \frac{\pi}{2} \right) = \pi \). Therefore, the original integral evaluates to \( \pi \).

Key Concepts

Limit of Integration

When dealing with improper integrals, it is important to understand why they are classified as such and how this impacts the calculation. In the exercise at hand, the integral \[ \int_{-2}^{2} \frac{d x}{\sqrt{4-x^{2}}} \] is improper because the integrand becomes undefined at both endpoints where the denominator hits zero. This becomes evident at - **Endpoints**: \( x = -2 \) and \( x = 2 \) - At these points, the expression under the square root, \( 4-x^2 \), becomes zero, leading to division by zero in the integrand \( \frac{1}{\sqrt{4-x^2}} \). Since the function cannot be evaluated at these points directly, the integral is defined as improper.To evaluate such an integral, we rely on the concept of limits: - **Expressing the Integral as Limits**: - This involves setting up the integral to approach these problematic points as limits: \[ \lim_{a \to 2^-} \int_{-2}^{a} \frac{1}{\sqrt{4-x^2}} \, dx + \lim_{b \to -2^+} \int_{b}^{2} \frac{1}{\sqrt{4-x^2}} \, dx \] - Here, we split the integration region to tackle the undefined points separately with limits. This ensures that we can evaluate the integral despite the division by zero.

Arcsine Function

The arcsine function plays a crucial role in solving this improper integral. Recognizing the form of the integrand as being similar to a derivative is key. The function \( \int \frac{1}{\sqrt{4-x^2}} \, dx \) resembles the derivative of the arcsine function. Specifically, the indefinite integral is \( \arcsin\left( \frac{x}{2} \right) + C \).This relationship hinges on the fact that:- **Derivative of the Arcsine Function**: - The derivative of \( \arcsin(\frac{x}{a}) \) is given by \( \frac{1}{\sqrt{a^2-x^2}} \), aligning perfectly with the form \( \sqrt{4-x^2} \) when \( a = 2 \). Through recognizing this connection, you can express the integrand in a form that allows integration using standard methods. This understanding is pivotal as it directly allows computing the indefinite integral that forms the core of our evaluation.

Evaluating Definite Integrals

Once the indefinite integral is known, evaluating the definite integrals follows naturally. The key is in correctly applying limits to the indefinite integral:- **Evaluating the Parts of the Integral**: - Before combining results, compute limits: - \[ \lim_{a \to 2^-} \left[ \arcsin\left(\frac{x}{2}\right) \right]_{-2}^{a} + \lim_{b \to -2^+} \left[ \arcsin\left(\frac{x}{2}\right) \right]_{b}^{2} \]. - **Limits and Definite Integration**: - Each component yields values deriving from the behavior of \( \arcsin \) at these critical points: * As \( a \to 2^- \), \( \arcsin(\frac{a}{2}) \to \frac{\pi}{2} \), and \( \arcsin(-1) = -\frac{\pi}{2} \). * As \( b \to -2^+ \), \( \arcsin(1) = \frac{\pi}{2} \), and \( \arcsin(\frac{b}{2}) \to -\frac{\pi}{2} \). - **Combining and Summing**: - Combine these: * \((\frac{\pi}{2} - (-\frac{\pi}{2})) + (\frac{\pi}{2} + \frac{\pi}{2}) = \pi\).Through this step-by-step evaluation, the entire integration process comes together revealing the result of \( \pi \), providing clarity on this once-improper integral.