Textbook Question
Explain the statement that a continuous function on an interval [a,b] equals its average value at some point on (a,b).
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Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.3.73
Derivatives of integrals Simplify the following expressions.
d/dπ β«βΛ£ (tΒ² + t + 1) dt
Verified step by step guidance1
Recognize that the problem involves the Fundamental Theorem of Calculus, which states that if F(x) = β«βΛ£ f(t) dt, then dF/dx = f(x).
Identify the integral bounds: the lower bound is a constant (3), and the upper bound is the variable (x). This means the derivative will be evaluated at the upper bound.
Apply the Fundamental Theorem of Calculus: d/dπ β«βΛ£ (tΒ² + t + 1) dt = (tΒ² + t + 1) evaluated at t = x.
Substitute t = x into the integrand: (xΒ² + x + 1).
Conclude that the derivative simplifies to xΒ² + x + 1, as the lower bound being constant does not contribute to the derivative.
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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
The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval, then the integral of f from a to b can be computed using F(b) - F(a). This theorem is crucial for simplifying expressions involving derivatives of integrals, as it allows us to evaluate the derivative of an integral directly.
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Fundamental Theorem of Calculus Part 1
Chain Rule
The Chain Rule is a fundamental principle in calculus used to differentiate composite functions. When differentiating an integral with variable limits, the Chain Rule helps in managing the relationship between the outer function (the derivative) and the inner function (the integral), ensuring that we account for changes in the variable of integration.
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05:02Intro to the Chain Rule
Definite Integral
A definite integral represents the accumulation of a function's values over a specific interval, providing a numerical result. In the context of the given expression, understanding how to evaluate the definite integral from 3 to x is essential for applying the Fundamental Theorem of Calculus and simplifying the expression correctly.
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05:43Definition of the Definite Integral
Related Practice
Textbook Question
Average values Find the average value of the following functions on the given interval. Draw a graph of the function and indicate the average value.
Ζ(π) = cos π on [βΟ/2 , Ο/2]
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Textbook Question
If Ζ is an odd function, why is β«α΅ββ Ζ(π) dπ = 0?
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Textbook Question
Mean Value Theorem for Integrals Find or approximate all points at which the given function equals its average value on the given interval.
Ζ(π) = 1 β |π| on [β1, 1]
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Textbook Question
Definite integrals Use geometry (not Riemann sums) to evaluate the following definite integrals. Sketch a graph of the integrand, show the region in question, and interpret your result.
β«ββΒ² ( β|π| ) dπ
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Textbook Question
Cubic zero net area Consider the graph of the cubic y = π (πβ a) (πβ b), where 0 < a < b. Verify that the graph bounds a region above the π-axis, for 0 < π < a , and bounds a region below the π-axis, for a < π < b. What is the relationship between a and b if the areas of these two regions are equal?
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