Question

Evaluate the following integrals or state that they diverge. $$\int_{-2}^{2} \frac{d x}{\sqrt{4-x^{2}}}$$

Short answer

Question: Evaluate the integral $$\int_{-2}^{2} \frac{1}{\sqrt{4-x^2}} dx$$. Answer: The integral $$\int_{-2}^{2} \frac{d x}{\sqrt{4-x^{2}}} = \frac{\pi}{2}$$.

Step-by-step solution

Identify the Function's Form

Notice that the given function is in the form $$\frac{1}{\sqrt{4-x^2}}$$, which resembles the formula for the arcsecant differentiation: $$\int \frac{1}{a \sqrt{1-\frac{x^2}{a^2}}} dx = \sec^{-1} (\frac{x}{a}) + C $$. In our case, a = 2.

Compute the Antiderivative

When the function is in the form of the arcsecant derivative, the antiderivative is as follows: $$\int \frac{1}{\sqrt{4-x^2}} dx = \frac{1}{2}\sec^{-1} (\frac{x}{2}) + C$$

Evaluate the Integral Using the Limits of Integration

Now that we have the antiderivative, we can evaluate the integral using the limits of integration -2 and 2: $$\int_{-2}^{2} \frac{1}{\sqrt{4-x^2}} dx = \frac{1}{2}\sec^{-1} (\frac{2}{2}) - \frac{1}{2}\sec^{-1} (\frac{-2}{2})$$

Simplify the Expression

Simplify the expression to get: $$\frac{1}{2}(\sec^{-1} (1) - \sec^{-1} (-1)) = \frac{1}{2}(\pi - 0) = \frac{\pi}{2}$$ #Answer# $$\int_{-2}^{2} \frac{d x}{\sqrt{4-x^{2}}} = \frac{\pi}{2}$$

Key concepts

Arcsecant Function

Understanding the arcsecant function is a central part of solving integrals that look like the one we're dealing with here. It is the inverse function of the secant function noted as \( \sec^{-1}(x) \). The secant function \( \sec(x) = \frac{1}{\cos(x)} \) is defined for angles, meaning its inverse, the arcsecant, gives us an angle when we know the secant of that angle.
The formula for arcsecant differentiation is important because it helps in finding antiderivatives of functions in specific forms. The arcsecant function applies in cases where the integral resembles \( \frac{1}{a \sqrt{1-\frac{x^2}{a^2}}} \), making it key in evaluating integrals of these kinds of expressions.

Divergent Integrals

Not all integrals can be evaluated to give a finite result. When an integral doesn't converge to a finite value, it is termed as divergent.
Here's what you should keep in mind about divergent integrals:

• If the limits of integration lead to a situation where the function grows unbounded.
• Often occur in improper integrals where the limits approach infinities or points where the function isn't defined.

For our integral, it doesn't diverge because it resolves to a specific value after applying arcsecant antiderivative, ensuring convergence to \( \frac{\pi}{2} \).

Antiderivative

Finding an antiderivative is necessary to solve definite integrals. An antiderivative is a function whose derivative equals the original function we want to integrate.
In the given exercise, the integral \( \frac{1}{\sqrt{4-x^2}} \) resembles a form suitable for obtaining its antiderivative using the arcsecant function:

• The identified antiderivative is \( \frac{1}{2}\sec^{-1}(\frac{x}{2}) + C \).
• This expression captures the inverse relationship of differentiation and integration using the arcsecant.

By evaluating this antiderivative over the given limits, we can compute the definite integral.

Limits of Integration

The concept of limits of integration is crucial when dealing with definite integrals. They help in specifying the exact range over which the function must be evaluated.
The definite integral is essentially the net area under the curve between two points, described by these limits. For instance, in our exercise:

• The integration limits are \(-2\) and \(2\).
• They determine what values we substitute into the antiderivative to obtain the final result: \( \frac{1}{2}(\sec^{-1} (1) - \sec^{-1} (-1)) = \frac{\pi}{2} \).

Understanding and applying these limits correctly are pivotal for solving the integral accurately, as shown in the final step of our exercise.