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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.6.64
Chapter 8, Problem 8.6.64

What is the largest value that
∫ from a to b x√(2x - x²) dx
can have for any a and b? Give reasons for your answer.

Verified step by step guidance
1
First, understand the problem: we want to find the largest possible value of the definite integral \(\int_a^b x \sqrt{2x - x^2} \, dx\) for any choice of limits \(a\) and \(b\). This means we are looking for the maximum area under the curve of the function \(f(x) = x \sqrt{2x - x^2}\) over some interval \([a,b]\) within the domain where the integrand is defined and real.
Determine the domain of the integrand \(f(x) = x \sqrt{2x - x^2}\). Since the expression inside the square root must be non-negative, solve \(2x - x^2 \geq 0\). Factor this as \(x(2 - x) \geq 0\), which implies \(x \in [0, 2]\). So the function is real-valued and defined on the interval \([0, 2]\).
To find the largest value of the integral, consider that the integral over any subinterval \([a,b] \subseteq [0,2]\) represents the area under the curve \(f(x)\). The maximum integral value will be the integral over the interval where the function is positive and the area is largest. Since the function is zero at the endpoints \(x=0\) and \(x=2\), and positive in between, the maximum integral is likely over the entire interval \([0,2]\).
Set up the integral over the full domain: \(I = \int_0^2 x \sqrt{2x - x^2} \, dx\). To solve this integral, use an appropriate substitution. For example, let \(u = 2x - x^2\), then find \(du\) in terms of \(dx\) and express \(x\) in terms of \(u\) to rewrite the integral in terms of \(u\). Alternatively, consider a trigonometric substitution to simplify the square root.
After substitution, rewrite the integral in a simpler form and evaluate it (this step involves integration techniques such as substitution or trigonometric substitution). The value of this integral over \([0,2]\) will give the largest possible value of the original integral for any \(a\) and \(b\) because any smaller interval will yield a smaller area.

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Key Concepts

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

Definite Integral and Area Interpretation

A definite integral ∫ from a to b f(x) dx represents the net area under the curve f(x) between x = a and x = b. Understanding this helps in visualizing how the integral's value changes with different limits a and b, especially when the integrand is positive or negative.
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Definition of the Definite Integral

Domain and Behavior of the Integrand

The integrand x√(2x - x²) is defined where the expression inside the square root, 2x - x², is non-negative. Identifying this domain (0 ≤ x ≤ 2) is crucial because the integral outside this range is not real-valued, and the function's shape within this domain determines where the integral attains its maximum.
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Maximizing the Integral by Choosing Limits

To find the largest value of the integral for any a and b, one must consider the integral over intervals where the integrand is positive and possibly use the Fundamental Theorem of Calculus. The maximum integral value occurs over the interval where the integrand accumulates the greatest positive area, typically between the points where the function is zero or changes sign.
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Example 3: Maximizing Profit
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