Textbook Question
2. When applying the formula for integration by parts, how do you choose the u and dv? How can you apply integration by parts to an integral of the form ∫ f(x) dx?
Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.AAE.26
Length of a curve
Find the length of the curve
y = ∫(from 1 to x) sqrt(sqrt(t) - 1) dt, where 1 ≤ x ≤ 16.
Verified step by step guidance1
Recognize that the curve is defined by an integral function: \(y = \int_{1}^{x} \sqrt{\sqrt{t} - 1} \, dt\). This means \(y\) is given as a function of \(x\) through an integral with variable upper limit.
Recall the formula for the length \(L\) of a curve \(y = f(x)\) from \(x = a\) to \(x = b\):
\(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\).
Use the Fundamental Theorem of Calculus to find \(\frac{dy}{dx}\). Since \(y = \int_{1}^{x} g(t) \, dt\) where \(g(t) = \sqrt{\sqrt{t} - 1}\), then
\(\frac{dy}{dx} = g(x) = \sqrt{\sqrt{x} - 1}\).
Substitute \(\frac{dy}{dx}\) into the length formula:
\(L = \int_{1}^{16} \sqrt{1 + \left(\sqrt{\sqrt{x} - 1}\right)^2} \, dx\).
Simplify the expression inside the square root and set up the integral for evaluation:
\(L = \int_{1}^{16} \sqrt{1 + (\sqrt{x} - 1)} \, dx = \int_{1}^{16} \sqrt{\sqrt{x}} \, dx\). This integral can then be evaluated using substitution or power rules.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definition of the Length of a Curve
The length of a curve y = f(x) from x = a to x = b is given by the integral of the square root of 1 plus the square of the derivative, ∫_a^b √(1 + (dy/dx)^2) dx. This formula measures the distance along the curve between two points.
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Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links differentiation and integration, stating that if y = ∫_a^x g(t) dt, then dy/dx = g(x). This allows us to find the derivative of an integral-defined function, which is essential for computing dy/dx in the curve length formula.
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Chain Rule and Simplification of Derivatives
When differentiating functions involving nested expressions, the chain rule helps find the derivative accurately. In this problem, simplifying dy/dx = sqrt(sqrt(x) - 1) is crucial before substituting into the length formula to evaluate the integral efficiently.
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