Textbook Question
In Exercises 75–78, find dy/dx.
y = ∫(from x to 1) (6/(3 + t^4))dt
Ch. 5 - Integrals
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 5, Problem 5.PE.35
Express the solutions of the initial value problems in Exercises 35 and 36 in terms of integrals.
dy/dx = sin x/x , y(5) = -3
Verified step by step guidance1
Identify the given differential equation and initial condition: \( \frac{dy}{dx} = \frac{\sin x}{x} \) with \( y(5) = -3 \).
Recognize that this is a first-order differential equation where \( \frac{dy}{dx} \) is given explicitly as a function of \( x \).
To find \( y(x) \), integrate both sides with respect to \( x \): \[ y(x) = \int \frac{\sin x}{x} \, dx + C \], where \( C \) is the constant of integration.
Use the initial condition \( y(5) = -3 \) to solve for \( C \): \[ -3 = \int_{}^{5} \frac{\sin t}{t} \, dt + C \]. Note that the variable of integration is changed to \( t \) to avoid confusion.
Express the solution explicitly in terms of an integral from 5 to \( x \): \[ y(x) = -3 + \int_5^{x} \frac{\sin t}{t} \, dt \]. This represents the solution in integral form.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Initial Value Problems (IVPs)
An initial value problem involves a differential equation along with a specified value of the unknown function at a given point. This initial condition allows us to find a unique solution that satisfies both the differential equation and the initial value.
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Initial Value Problems
Separable Differential Equations and Integration
When a differential equation is given in the form dy/dx = f(x), the solution can be found by integrating f(x) with respect to x. This process involves expressing y as an integral of the function plus a constant determined by the initial condition.
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Definite Integrals and Applying Initial Conditions
To express the solution in terms of integrals, we use definite integrals with limits corresponding to the initial condition. This approach incorporates the initial value directly, allowing the constant of integration to be evaluated and the solution to be written explicitly.
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Related Practice
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lim ∑ (2cₖ - 1)⁻¹/² ∆xₖ, where P is a partition of [1, 5]
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