Problem 74 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{2}^{4} \frac{2 d x}{x^{2}-6 x+10}$$

Step-by-Step Solution

Answer

The value of the integral is \( \pi \).

Step 1: Identify the form of the integral

The given integral is \( \int_{2}^{4} \frac{2 \, dx}{x^{2}-6x+10} \) which corresponds to the form \( \int \frac{1}{(ax - b)^2 + c^2} \, dx \). We need to rewrite the denominator to fit this form.

Step 2: Complete the square in the denominator

Rewrite the quadratic expression in the denominator by completing the square:\[x^2 - 6x + 10 = (x-3)^2 + 1\]Thus, the integral becomes \( \int_{2}^{4} \frac{2 \, dx}{(x-3)^2 + 1^2} \).

Step 3: Identify the standard integral form

Recognize that the integral now matches the standard form \( \int \frac{k \, dx}{(x-a)^2 + b^2} = \frac{k}{b} \arctan\left(\frac{x-a}{b}\right) + C \), where \( k = 2 \), \( a = 3 \), and \( b = 1 \).

Step 4: Apply the integral formula

Apply the formula:\[\int \frac{2 \, dx}{(x-3)^2 + 1} = 2 \arctan(x-3) + C\]

Step 5: Evaluate the integral with limits

Substitute the limits into the evaluated integral:\[\begin{aligned}F(x) &= 2 \arctan(x-3) \Big|_2^4 \&= 2 \arctan(4-3) - 2 \arctan(2-3)\end{aligned}\]Calculate further:\[= 2 \arctan(1) - 2 \arctan(-1)\]

Step 6: Substitute known values of arctan

Utilize known values: \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(-1) = -\frac{\pi}{4} \):\[= 2 \left( \frac{\pi}{4} \right) - 2 \left( -\frac{\pi}{4} \right)\]

Step 7: Simplify to find the solution

Simplify the expression:\[= 2 \cdot \frac{\pi}{4} + 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} + \frac{\pi}{2} = \pi\]

Key Concepts

Integral Calculus

Integral calculus is a vital branch of mathematics that deals with integrals, which are essentially mathematical tools used to accumulate quantities, such as areas under curves. To find an integral, the process usually involves reversing differentiation. In the given exercise, we deal with a definite integral, meaning we are finding the area under the curve defined by the function between the specified limits, 2 and 4.

• An integral is represented as \( \int f(x) \, dx \), where \( f(x) \) is the integrand.
• Definite integrals have limits, so you calculate a specific value that corresponds to the area under the curve.
• Indefinite integrals, however, don’t have limits and typically result in the function plus a constant.

In our case, the integral involves a rational function that simplifies using methods like completing the square. The solution requires substituting known values, establishing a relationship with standard integral forms for an easier evaluation, as seen later with the arctangent function.

Arctangent Function

The arctangent function is an inverse trigonometric function, denoted as \( \arctan(x) \). This function essentially gives the angle whose tangent is the number \( x \). In calculus, it appears frequently in integrals, especially when dealing with expressions in the form \( \frac{1}{x^2 + 1} \) or similar, which result in the arctangent function when integrated.

• For example, \( \int \frac{1}{1 + x^2} \, dx = \arctan(x) + C \).
• The arctangent is useful because it maps any real number to an angle between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).

In the exercise, the integration yielded \( 2 \arctan(x-3) \). The constants and linear transformations (introduced by completing the square) are adjustments to fit the arctangent's pattern. This provides a straightforward application of the arctangent integral formula during calculations.

Completing the Square

Completing the square is a powerful algebraic method used to transform quadratic expressions into a perfect square form. This is highly beneficial in calculus for easily evaluating integrals involving quadratic expressions. The process takes a quadratic form \( ax^2 + bx + c \) and rewrites it as \( (x - h)^2 + k \), making the execution of subsequent calculations more manageable.

• To complete the square, you take the coefficient of \( x \), halve it, and then square it.
• Add and subtract that squared value within the expression to maintain balance.
• The goal is to bring the quadratic into the form \( (x-a)^2 + b^2 \), which is simpler for integration.

In this exercise, the quadratic \( x^2 - 6x + 10 \) is rewritten as \( (x-3)^2 + 1 \). This neat transformation fits comfortably into the integral form related to the arctangent function. As a result, it reduces a complex expression into something easier to handle, enabling a direct application of integral formulas.