Problem 38 - Thomas Calculus

Question

Evaluate each integral in Exercises \(37-42\) by completing the square and using a substitution to reduce it to standard form. $$ \int_{2}^{4} \frac{2 d x}{x^{2}-6 x+10} $$

Step-by-Step Solution

Answer

The integral evaluates to \(\pi\).

Step 1: Rewrite the Quadratic

The integrand contains the quadratic expression \(x^2 - 6x + 10\). We begin by completing the square for this quadratic. First, take the coefficient of \(x\), which is 6, divide it by 2, giving 3, and then square it to get 9. Thus, \(x^2 - 6x + 10 = (x-3)^2 + 1\).

Step 2: Set Up the Substitution

Let's use the substitution \(u = x - 3\), which gives \(du = dx\). The limits of integration change accordingly: when \(x = 2\), \(u = 2 - 3 = -1\), and when \(x = 4\), \(u = 4 - 3 = 1\).

Step 3: Substitute and Simplify the Integral

Substitute \(x = u + 3\) into the integral. The integral becomes \(\int_{-1}^{1} \frac{2}{u^2 + 1} du\). This is a standard integral form which corresponds to \( \int \frac{1}{u^2 + 1} du = \arctan(u)\).

Step 4: Integrate and Evaluate

The integral from the previous step is \(2 \int_{-1}^{1} \frac{1}{u^2 + 1} du = 2[\arctan(u)]_{-1}^{1}\). Evaluating this gives \(2(\arctan(1) - \arctan(-1))\).

Step 5: Calculate the Arctan Values

The function \(\arctan(1)\) is \(\frac{\pi}{4}\) and \(\arctan(-1)\) is \(-\frac{\pi}{4}\). Substituting these values, we have \(2\left(\frac{\pi}{4} - (-\frac{\pi}{4})\right) = 2\times \frac{\pi}{2} = \pi\).

Key Concepts

Completing the Square

Completing the square is a useful technique for simplifying expressions, particularly quadratics, into a form that's easier to work with in calculus. When you see a quadratic expression like \(x^2 - 6x + 10\), you might notice it's in standard form. However, to make integration or other operations easier, we can rewrite it by completing the square.
Here's how it's done:

• Identify the coefficient of \(x\), which in this case is \(-6\).
• Divide this coefficient by 2, resulting in \(-3\).
• Square this result to get \(9\).
• Now, rewrite the quadratic as a perfect square trinomial. Thus, \(x^2 - 6x + 10 = (x-3)^2 + 1\).
  This turns the expression into a perfect square plus a constant, greatly simplifying the process of integration or other calculus operations.

This approach is quite powerful in problems involving integrals with quadratics because it sets up further steps in a more standardized format, making our work easier.

Definite Integral

Definite integrals are a central part of integral calculus, providing the total accumulation of a quantity between two points. In this context, we're looking at the integral \(\int_{2}^{4} \frac{2 d x}{x^{2}-6 x+10}\).
Definite integrals are calculated between specific limits—a lower and an upper bound. They help calculate areas under a curve or other accumulated changes such as distance:

• The bounds here are from 2 to 4, indicating we're interested in the area under the curve from \(x = 2\) to \(x = 4\).
• Definite integrals involve both evaluating the antiderivative and applying the limits of integration to find the accumulated value.

After performing operations such as substitution, we solve the integral within these limits and use the fundamental theorem of calculus to determine the total value accumulated over this interval, simplifying and translating into real-world problems or further mathematical manipulation.

Trigonometric Substitution

Trigonometric substitution is a powerful technique employed when dealing with integrals that contain specific types of quadratic expressions. It succeeds by transforming an unwieldy algebraic expression into an easier trigonometric form.

• In the exercise, after completing the square, we have \((x-3)^2 + 1\). This suggests a link to the identity \(1 + u^2\) which is tied to trigonometric functions.
• Performing the substitution \(u = x - 3\), so that \(du = dx\), simplifies the original integral to a well-known trigonometric form \(\int \frac{1}{u^2 + 1} du\).

In this specific context, this substitution aids in rewriting the integral in a way that is directly tied to the arctan function, which is inherently trigonometric. This transformation concentrates complex algebra into a straightforward integration task that uses standard trigonometric results.

Arctangent Function

The arctangent function, or \(\arctan\), is crucial in solving integrals that involve quadratics turned into the form of \(\frac{1}{x^2 + 1}\). As a standard integral form, knowing \(\arctan\) is helpful when simplifying problems using trigonometric substitution.

• When you integrate \(\frac{1}{u^2 + 1}\), the result is \(\arctan(u) + C\), where \(C\) represents the constant of integration for indefinite integrals.
• In a definite integral, this translates into evaluating \(\arctan(u)\) at the upper and lower bounds, as was done in this exercise from \(-1\) to \(1\).
• The values \(\arctan(1) = \frac{\pi}{4}\) and \(\arctan(-1) = -\frac{\pi}{4}\) demonstrate how \(\arctan\) maps real numbers to angles, consolidating the integral's result due to properties of angles and circular functions.

Thus, having a firm understanding of \(\arctan\) assists in recognizing standard integration forms and their applications across varying calculus problems.