Question

Evaluate the following integrals. $$\int_{-2}^{-1} \sqrt{-4 x-x^{2}} d x$$

Short answer

In this problem, we first analyzed the integrand and found that the quadratic expression is non-negative within the given interval. Then, we found the primitive function using substitution and trigonometric substitution methods. Finally, we applied the fundamental theorem of calculus to evaluate the definite integral with the given limits. The solution to the integral is: $$\int_{-2}^{-1} \sqrt{-4 x-x^{2}} d x=-\frac{\pi}{2}$$

Step-by-step solution

Investigate the integrand

Before diving into integration, let's examine the integrand: $$f(x) = \sqrt{-4 x - x^2}$$ The expression inside the square root is a quadratic function of the form \(ax^2 + bx + c\), where \(a = -1\), \(b = -4\), and \(c = 0\). To guarantee the integrand is a real number for all values of x within the interval, the expression inside the square root should be non-negative. In other words, the quadratic must be non-negative for \(-2 \le x \le -1\). Let's analyze the quadratic function to confirm this about the domain. The quadratic opens downwards since its leading coefficient is negative, meaning it forms a parabola with a highest point that is non-negative in the given interval.

Find the primitive

Now, let's find an antiderivative or primitive of the given integrand. This antiderivative is a function that represents the area under the curve of the integrand up to some point x in the interval, and when differentiated, results in the original integrand. Finding the primitive of such functions might be challenging. In our case, we'll use a substitution method. Observe that: $$-4x - x^2 = 4(-x - \frac{x^2}{4}) = 4(1 - (x + 1)^2)$$ Then, let's make a substitution: $$u = x + 1 \implies du = dx$$ The integrand becomes: $$f(x) = \sqrt{4(1 - u^2)} = 2\sqrt{1 - u^2}$$ Thus, we can rewrite the integral as: $$\int_{-1}^{0} 2\sqrt{1 - u^2} du$$ Now, it becomes clear that we are dealing with an integral of the form \(\int \sqrt{1 - u^2} du\), which is a well-known integral that can be evaluated using a trigonometric substitution: Let \(u = \sin{v} \implies du = \cos{v} dv\). Then, $$\int 2\sqrt{1 - u^2} du = 2 \int \sqrt{1 - \sin^2{v}} \cos{v} dv = 2 \int \cos^2{v} dv$$ Now, we can apply the power-reduction formula to evaluate the integral: $$\cos^2{v} = \frac{1 + \cos{2v}}{2}$$ Substituting this back into the integral, we get: $$2 \int \cos^2{v} dv = 2 \int \frac{1 + \cos{2v}}{2} dv = \int (1 + \cos{2v}) dv$$ Now, we can integrate term by term: $$\int (1 + \cos{2v}) dv = v + \frac{1}{2} \sin{2v} + C$$ Now, we undo the substitutions we made earlier: $$v + \frac{1}{2} \sin{2v} + C = \arcsin{(x+1)} + \frac{1}{2} \sin{(2\arcsin{(x+1)})} + C$$ This is the primitive function, F(x), we were looking for.

Evaluate the definite integral

Now, we apply the fundamental theorem of calculus to evaluate the definite integral: $$\int_{-2}^{-1} \sqrt{-4 x-x^{2}} d x = F(-1) - F(-2) = (\arcsin{0} + \frac{1}{2} \sin{0}) - (\arcsin{1} + \frac{1}{2} \sin{\pi})$$ Simplifying the expression, we get: $$0 - (\frac{\pi}{2} + 0) = -\frac{\pi}{2}$$ So, the solution to the given integral is: $$\int_{-2}^{-1} \sqrt{-4 x-x^{2}} d x=-\frac{\pi}{2}$$

Key concepts

What is an Antiderivative?

An antiderivative, also known as an indefinite integral or primitive, is a function that reverses the process of differentiation. In simpler terms, if you have a function f(x), an antiderivative is another function F(x) such that when F(x) is differentiated, it yields f(x). That is, F'(x) = f(x).

Consider an everyday analogy: if you were walking at a constant speed (let's say, speed being the derivative of your position), your position as a function of time would be the antiderivative of your speed.

In the context of integrals, the antiderivative plays a crucial role as it helps in evaluating definite integrals over an interval [a, b]. According to the Fundamental Theorem of Calculus, if F(x) is an antiderivative of f(x), the definite integral of f(x) from a to b is F(b) - F(a). Simplifying a complicated function such as f(x) = \(\sqrt{-4x - x^2}\) to its antiderivative can be challenging. However, by using techniques like substitution, this process can be made more accessible.

The Substitution Method in Calculus

The substitution method is a technique in integral calculus that simplifies the integration process by changing the variable of integration to another, more convenient variable. This method is useful particularly when dealing with complex functions whose antiderivatives are not immediately obvious.

To perform a substitution, you choose a new variable u, which is a function of x, and express the integrand in terms of u. By finding du, which is the derivative of u with respect to x, you can replace dx in the original integral. This transformation simplifies the integral, often to a form that is easier to evaluate.

As part of the exercise improvement advice, it's essential to clarify the substitution choice. In the given exercise, the substitution u = x + 1 was cleverly chosen to transform a problematic square root containing a quadratic expression into a simple expression involving u. This leads to a form that allows the use of standard integral solutions or further simplification through additional methods, like trigonometric substitution.

Trigonometric Substitution

When an integrand involves the square root of a quadratic expression, like \(\sqrt{1 - u^2}\), trigonometric substitution is a powerful technique to simplify the integral. This method involves substituting variables in the integrand with trigonometric functions that correspond to identities or properties found in the Pythagorean trigonometric identity: sin^2(θ) + cos^2(θ) = 1.

The choice of trigonometric function depends on the form of the quadratic expression, but for \(\sqrt{1 - u^2}\), the substitution u = sin(v) is appropriate because it leads directly to a recognizable trigonometric identity. After this substitution, the differential du becomes \(cos(v) dv\), and the integral turns into an expression in terms of v that is much easier to integrate.

In the definite integral exercise provided, trigonometric substitution simplifies the integrand to \(2\cos^2(v)\), which then allows the use of the power-reduction formula to obtain an integral with straightforward antiderivatives. This method showcases the combining of different techniques - substitution followed by trigonometric substitution - to evaluate a definite integral that at first glance seems intimidating.