Textbook Question
Determine whether the following statements are true and give an explanation or counterexample.
b. The area of the region between y=sin x and y=cos x on the interval [0,Ο/2] is β«Ο/20(cosxβsinx)dx.
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9. Graphical Applications of Integrals
Area Between Curves
4:29 minutesProblem 5.5.96
Textbook Question
Areas of regions Find the area of the following regions.
The region bounded by the graph of Ζ(π) = x /β(πΒ² β9) and the π-axis between and π = 4 and π= 5
Verified step by step guidance1
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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Video duration:
4mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 / β(xΒ² - 9) is crucial because the expression under the square root must be positive, so xΒ² - 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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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 / β(xΒ² - 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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