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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.5.96
Capitolo 5, Problema 5.5.96

Areas of regions Find the area of the following regions.                                                                                                                   
                                                                                                                                                                 The region bounded by the graph of ƒ(𝓍) = x /√(𝓍² ―9) and the 𝓍-axis between and 𝓍 = 4 and 𝓍= 5

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Identify the function and the interval: The function given is \(f(x) = \frac{x}{\sqrt{x^{2} - 9}}\), and the region is bounded by this curve and the x-axis between \(x=4\) and \(x=5\).
Determine the points where the function intersects the x-axis: Since the function involves a square root in the denominator, check the domain. The expression under the square root, \(x^{2} - 9\), must be positive, so \(|x| > 3\). The interval \([4,5]\) is valid. Also, note that the function is positive or negative in this interval to understand the area calculation.
Set up the definite integral for the area: The area between the curve and the x-axis from \(x=4\) to \(x=5\) is given by the integral \(A = \int_{4}^{5} \left| \frac{x}{\sqrt{x^{2} - 9}} \right| \, dx\). Since the function is positive in this interval, the absolute value can be removed.
Simplify the integral if possible: Consider a substitution to evaluate the integral. For example, let \(u = x^{2} - 9\), then \(du = 2x \, dx\), which can help rewrite the integral in terms of \(u\).
Express the integral in terms of \(u\) and set the new limits: When \(x=4\), \(u = 4^{2} - 9 = 16 - 9 = 7\), and when \(x=5\), \(u = 25 - 9 = 16\). Rewrite the integral accordingly and prepare to integrate with respect to \(u\).

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Definite Integral for Area Calculation

The definite integral of a function between two points gives the net area between the curve and the x-axis over that interval. When the function is positive, the integral represents the area under the curve. To find the area of a region bounded by a curve and the x-axis, you evaluate the definite integral of the function from the lower to the upper limit.
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Definition of the Definite Integral

Domain and Behavior of the Function

Understanding the domain of the function f(x) = x / √(x² - 9) is crucial because the expression under the square root must be positive, so x² - 9 > 0, implying |x| > 3. This ensures the function is real-valued and continuous on the interval [4,5]. Recognizing this helps avoid invalid integration limits and ensures the integral is well-defined.
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Finding the Domain and Range of a Graph

Handling Functions with Square Roots in the Denominator

Functions with square roots in the denominator often require algebraic manipulation or substitution to integrate effectively. For f(x) = x / √(x² - 9), a common technique is to use a trigonometric or hyperbolic substitution to simplify the integral. This approach transforms the integral into a more manageable form, facilitating the calculation of the area.
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Limits of Rational Functions: Denominator = 0
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Max/min of area functions Suppose ƒ is continuous on [0 ,∞) and A(𝓍) is the net area of the region bounded by the graph of ƒ and the t-axis on [0, x]. Show that the local maxima and minima of A occur at the zeros of ƒ. Verify this fact with the function ƒ(𝓍) = 𝓍² - 10𝓍.

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Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

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