Textbook Question
Evaluate the integrals in Exercises 47–68.
∫₁² 4 dv
v²
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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.77
In Exercises 75–78, find dy/dx.
y = ∫(from x to 1) (6/(3 + t^4))dt
Verified step by step guidance1
Recognize that the function y is defined as a definite integral with a variable limit: \(y = \int_{x}^{1} \frac{6}{3 + t^{4}} \, dt\).
Recall the Leibniz rule for differentiation of an integral with variable limits: if \(y = \int_{a(x)}^{b(x)} f(t) \, dt\), then \(\frac{dy}{dx} = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)\).
Identify the limits of integration: the lower limit is \(x\) and the upper limit is the constant \(1\). Therefore, \(a(x) = x\) and \(b(x) = 1\).
Calculate the derivatives of the limits: \(a'(x) = \frac{d}{dx} x = 1\) and \(b'(x) = \frac{d}{dx} 1 = 0\).
Apply the Leibniz rule: \(\frac{dy}{dx} = f(1) \cdot 0 - f(x) \cdot 1 = - \frac{6}{3 + x^{4}}\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Fundamental Theorem of Calculus Part 1
This theorem connects differentiation and integration, stating that if a function is defined as an integral with a variable limit, its derivative is the integrand evaluated at that limit, multiplied by the derivative of the limit. It allows us to differentiate integrals with variable limits directly.
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Fundamental Theorem of Calculus Part 1
Leibniz Rule for Differentiation under the Integral Sign
Leibniz Rule generalizes differentiation of integrals with variable limits. When both limits depend on x, the derivative of the integral is the integrand evaluated at the upper limit times the derivative of the upper limit minus the integrand at the lower limit times the derivative of the lower limit.
Recommended video:
05:56Additional Rules for Indefinite Integrals
Chain Rule
The chain rule is used to differentiate composite functions. When the limits of integration are functions of x, the derivative of the integral involves multiplying the integrand evaluated at the limit by the derivative of that limit, applying the chain rule to handle the inner function.
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05:02Intro to the Chain Rule
Related Practice
Textbook Question
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
Textbook Question
If ∫₀² ƒ(x) dx = π, ∫₀² 7g(x) dx = 7, and ∫₀¹ g(x) dx = 2, find the value of each of the following.
_
d. ∫₀² √2ƒ(x) dx
Textbook Question
If ∫²₋₂ 3ƒ(x) dx = 12, ∫⁵₋₂ ƒ(x) dx = 6, and ∫⁵₋₂ g(x) dx = 2, find the value of each of the following.
d. ∫⁵₋₂ (-πg(x)) dx
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Textbook Question
Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.
√x + √y = 1, x = 0, y = 0
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Textbook Question
Definite Integrals
In Exercises 5–8, express each limit as a definite integral. Then evaluate the integral to find the value of the limit. In each case, P is a partition of the given interval, and the numbers cₖ are chosen from the subintervals of P.
n
lim ∑ (cos(cₖ/2)) ∆xₖ, where P is a partition of [-π, 0]
∥P∥→0 k = 1
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