Problem 77 - Calculus Early Transcendentals

Question

Evaluate the following integrals. Consider completing the square. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$

Step-by-Step Solution

Answer

**Answer:** The integral is equal to: $$\sinh^{-1}(2) - \sinh^{-1}(\sqrt{2})$$

Step 1: Completing the square for the expression inside the square root

To complete the square, we want to rewrite the expression \((x-1)(x-3)\) as a perfect square. We can start by expanding the expression: \((x-1)(x-3) = x^2-4x+3\) Next, add and subtract the square of half of the coefficient of the x-term (in this case, \((-4)/2 = -2\)): \(x^2-4x+3 = (x^2-4x+4)-1 = (x-2)^2 - 1\) Now, we can replace the expression inside the integral with the newly found expression: $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-2)^2 - 1}}$$

Step 2: Perform substitution

Let \(u=x-2\). Then \(du = dx\), and the limits of integration become \(u = \sqrt{2}\) (when \(x = 2 + \sqrt{2}\)) and \(u = 2\) (when \(x = 4\)). We rewrite the integral as: $$\int_{\sqrt{2}}^{2} \frac{d u}{\sqrt{u^2 - 1}}$$

Step 3: Recognize the hyperbolic trigonometric function

Notice that the integrand is similar to the derivative of inverse hyperbolic sine function, which is given by: $$\frac{d}{du} \sinh^{-1}(u) = \frac{1}{\sqrt{u^2 - 1}}$$ As the integrand matches the derivative, we can directly write the antiderivative: $$\int \frac{d u}{\sqrt{u^2 - 1}} = \sinh^{-1}(u) + C$$

Step 4: Evaluate the integral using the Fundamental Theorem of Calculus

Now, we just have to compute the difference between the antiderivative evaluated at the limits of integration to find the value of the integral: $$\int_{\sqrt{2}}^{2} \frac{d u}{\sqrt{u^2 - 1}} = \left[\sinh^{-1}(u) \right]_{\sqrt{2}}^{2} = \sinh^{-1}(2) - \sinh^{-1}(\sqrt{2})$$

Key Concepts

Completing the Square

Completing the square is a useful algebraic technique that allows us to rewrite quadratic expressions in a form that is easier to manipulate, especially for integration. When dealing with integrals involving square roots, completing the square transforms the expression into a perfect square, simplifying the evaluation of the integral.

To complete the square for a quadratic expression of the form \[ax^2 + bx + c\]we need to transform it into \[(x-h)^2 + k\]where \(h\) and \(k\) are constants. Let's break it down into a step-by-step process:

• First, factor out the coefficient of \(x^2\) if it isn't 1. In our case with \[x^2 - 4x + 3\], the leading coefficient is already 1.
• Next, take half of the \(x\)-term coefficient (\(-4/2 = -2\)) and square it to get 4.
• Add and subtract this square inside the expression: \[x^2 - 4x + 3 = (x^2 - 4x + 4) - 1 = (x-2)^2 - 1\].

This technique is widely applicable not only in algebraic expressions but also in solving quadratic equations, integrating functions, and analyzing functions' behavior.

Hyperbolic Functions

Hyperbolic functions, such as hyperbolic sine (\(\sinh\)), can be likened to the more familiar trigonometric functions, but with distinct properties that make them especially useful in calculus and complex number analysis. Each hyperbolic function corresponds to a trigonometric function:

• Hyperbolic sine: \(\sinh(x)\)
• Hyperbolic cosine: \(\cosh(x)\)
• Hyperbolic tangent: \(\tanh(x)\)

In the context of the integral given in the exercise, recognizing the expression inside the integral as matching the derivative of an inverse hyperbolic function allows us to integrate directly. For example, \[\frac{d}{du} \sinh^{-1}(u) = \frac{1}{\sqrt{u^2 - 1}}\]makes it possible to recognize the integral of \[\frac{1}{\sqrt{u^2-1}}\] as \[\sinh^{-1}(u)\]. Clear understanding of these derivatives makes tackling integrals involving hyperbolic functions more straightforward.

These hyperbolic functions have profound applications across different areas, including hyperbolic geometry, special relativity, and mathematical modeling of hyperbolic shapes.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of integral calculus, providing a connection between differentiation and integration. It states that if a function is continuous over an interval, an integral of the function over that interval can be determined using its antiderivative. Essentially, the theorem has two parts:

• Part 1: Provides a way to construct an antiderivative from a given integral. If \(f(x)\) is continuous on \([a, b]\), then the function \(F\) defined by \[F(x) = \int_a^x f(t) \, dt\]is an antiderivative of \(f\).
• Part 2: States that if \(F\) is an antiderivative of \(f\) on the interval \([a, b]\), the definite integral of \(f\) from \(a\) to \(b\) is given by \[\int_a^b f(x) \, dx = F(b) - F(a)\].

In practice, this means that once an antiderivative of the function is known, we can easily evaluate the integral by finding the values of the antiderivative at the boundaries of the integral and subtracting them.

In the exercise, after expressing the integrand as a known antiderivative involving \(\sinh^{-1}(u)\), the Fundamental Theorem of Calculus is applied to evaluate the integral from \(\sqrt{2}\) to 2 by calculating \(\sinh^{-1}(2) - \sinh^{-1}(\sqrt{2})\). This highlights the power of the theorem in transforming potential complex computations into straightforward evaluations.