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In Exercises 75–78, find dy/dx.
y = ∫(from x to 1) (6/(3 + t^4))dt
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Ch. 5 - Integrals
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- 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 Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 5, Problema 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
Guida verificata passo dopo passo1
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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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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05:03
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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05:43Definition of the Definite Integral
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