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.9d
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
Verified step by step guidance1
Recall the property of definite integrals that allows constants to be factored out: for any constant \(c\), \(\int_a^b c \cdot h(x) \, dx = c \int_a^b h(x) \, dx\).
Apply this property to the integral \(\int_{-2}^5 (-\pi g(x)) \, dx\), rewriting it as \(-\pi \int_{-2}^5 g(x) \, dx\).
Use the given value \(\int_{-2}^5 g(x) \, dx = 2\) and substitute it into the expression to get \(-\pi \times 2\).
Express the integral in terms of \(\pi\) and the known value without calculating the numerical product, as per instructions.
Thus, the value of \(\int_{-2}^5 (-\pi g(x)) \, dx\) is \(-2\pi\) times the integral of \(g(x)\) over \([-2,5]\), which is \(-2\pi\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Properties of Definite Integrals
Definite integrals have linearity properties, meaning the integral of a sum is the sum of the integrals, and constants can be factored out. For example, ∫[a to b] c·f(x) dx = c·∫[a to b] f(x) dx. This allows simplification of integrals involving constants or sums of functions.
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Definition of the Definite Integral
Given Integral Values and Interval Consistency
When solving problems involving definite integrals, it is crucial to pay attention to the integration limits. The given values correspond to specific intervals, and the integral's value depends on these limits. Ensuring the limits match the problem's integral is essential for correct evaluation.
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Integral of a Scalar Multiple of a Function
If a function g(x) is multiplied by a scalar constant k, the integral over an interval is k times the integral of g(x) over that interval. Formally, ∫[a to b] k·g(x) dx = k·∫[a to b] g(x) dx. This property simplifies calculations involving constants like -π.
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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
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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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.
b. ∫₁² g(x) dx
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