7–84. Evaluate the following integrals.
9. ∫ from 4 to 6 [1 / √(8x – x²)] dx

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Recognize that the integral ∫ from 4 to 6 [1 / √(8x – x²)] dx involves a square root in the denominator, which suggests a trigonometric substitution might simplify the expression. Specifically, the term 8x - x² can be rewritten in a standard quadratic form.

Rewrite 8x - x² as - (x² - 8x). Complete the square for the quadratic expression x² - 8x by adding and subtracting 16: x² - 8x = (x - 4)² - 16. Thus, 8x - x² becomes 16 - (x - 4)².

Substitute x - 4 = 4sin(θ), which implies x = 4 + 4sin(θ). This substitution transforms the square root √(16 - (x - 4)²) into a trigonometric expression. Also, compute dx = 4cos(θ)dθ.

Change the limits of integration based on the substitution. When x = 4, θ = arcsin(0) = 0. When x = 6, θ = arcsin(1/2). Update the integral accordingly.

Substitute the expressions for x, dx, and the square root into the integral. Simplify the resulting trigonometric integral and evaluate it over the new limits. Use standard trigonometric identities and integration techniques to complete the solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals

A definite integral represents the signed area under a curve between two specified limits. It is denoted as ∫ from a to b f(x) dx, where 'a' and 'b' are the lower and upper limits, respectively. The result of a definite integral is a numerical value that quantifies the accumulation of the function's values over the interval [a, b].

Integration Techniques

Integration techniques are methods used to evaluate integrals that may not be solvable by basic antiderivatives. Common techniques include substitution, integration by parts, and trigonometric substitution. For the integral in the question, recognizing the form of the integrand may suggest a suitable technique to simplify the evaluation process.

Improper Integrals

An improper integral occurs when the integrand has an infinite discontinuity or when the limits of integration are infinite. In this case, the integral must be evaluated as a limit. Understanding how to handle improper integrals is crucial for determining convergence and calculating the area under curves that may not be well-defined at certain points.

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