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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 5 - IntegrationProblem 5.4.58a
Chapter 5, Problem 5.4.58a

Bounds on an integral Suppose ƒ is continuous on [a, b] with ƒ''(𝓍) > 0 on the interval. It can be shown that (b―a) ƒ [(a + b) /2] ≤ ∫ₐᵇ ƒ(𝓍) d𝓍 ≤ (b―a) [ (ƒ(a) + ƒ(b)) /2]                                                         
                                                                                                                                                                               
(a) Assuming ƒ is nonnegative on [a, b], draw a figure to illustrate the geometric meaning of these inequalities. Discuss your conclusions. b. 

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Step 1: Understand the given inequality. Since \( f''(x) > 0 \) on \([a, b]\), the function \( f \) is convex on this interval. The inequality states that: \[ (b - a) f\left(\frac{a + b}{2}\right) \leq \int_a^b f(x) \, dx \leq (b - a) \frac{f(a) + f(b)}{2} \] This means the integral of \( f \) over \([a, b]\) is bounded below by the area of a rectangle with height \( f \) at the midpoint, and above by the area of a trapezoid formed by the values at the endpoints.
Step 2: Sketch the graph of \( f \) on the interval \([a, b]\). Since \( f \) is convex, the curve lies below the line segment connecting \( (a, f(a)) \) and \( (b, f(b)) \). Mark the points \( (a, f(a)) \), \( (b, f(b)) \), and the midpoint \( \left(\frac{a+b}{2}, f\left(\frac{a+b}{2}\right)\right) \).
Step 3: Illustrate the lower bound: draw a rectangle with base \( (b - a) \) and height \( f\left(\frac{a+b}{2}\right) \). This rectangle represents the quantity \( (b - a) f\left(\frac{a+b}{2}\right) \), which is less than or equal to the integral.
Step 4: Illustrate the upper bound: draw the trapezoid formed by the points \( (a, 0) \), \( (a, f(a)) \), \( (b, f(b)) \), and \( (b, 0) \). The area of this trapezoid is \( (b - a) \frac{f(a) + f(b)}{2} \), which is greater than or equal to the integral.
Step 5: Discuss conclusions: Since \( f \) is convex and nonnegative, the integral (area under the curve) lies between the area of the midpoint rectangle and the trapezoid formed by the endpoints. This reflects how convexity controls the shape of \( f \) and provides useful bounds for the integral.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Convexity and the Second Derivative Test

A function ƒ is convex on [a, b] if its second derivative ƒ''(x) is positive throughout the interval. This means the graph of ƒ lies below any chord connecting two points on the curve. Convexity ensures certain inequalities hold for integrals and averages of the function values.
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Integral Bounds and the Midpoint and Trapezoidal Approximations

The integral of ƒ over [a, b] can be bounded using the midpoint rule (evaluating ƒ at (a+b)/2) and the trapezoidal rule (averaging ƒ(a) and ƒ(b)). For convex functions, the midpoint approximation underestimates the integral, while the trapezoidal approximation overestimates it, providing useful bounds.
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Geometric Interpretation of Integral Inequalities

The inequalities relate areas under the curve to areas of simple geometric shapes: a rectangle at the midpoint and a trapezoid formed by endpoints. For a convex function, the curve lies below the trapezoid and above the midpoint rectangle, visually explaining why the integral is bounded between these two approximations.
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Related Practice
Textbook Question

The velocity in ft/s of an object moving along a line is given by v = ƒ(t) on the interval 0 ≤ t ≤ 8 (see figure), where t is measured in seconds.

a) Divide the interval [0,8] into n = 2 subintervals, [0,4] and [4,8]. On each subinterval, assume the object moves at a constant velocity equal to the value of v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0,8] (see part (a) of the figure)                                                                                                             


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Textbook Question

Working with area functions Consider the function ƒ and the points a, b, and c.

(a) Find the area function A (𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem.

ƒ(𝓍) = ― 12𝓍 (𝓍―1) (𝓍― 2) ; a = 0 , b = 1 , c = 2

Textbook Question

Average value with a parameter Consider the function ƒ(𝓍) = a𝓍 (1―𝓍) on the interval [0, 1], where a is a positive real number.

(a) Find the average value of ƒ as a function of a .

Textbook Question

Use Table 5.6 to evaluate the following indefinite integrals.                                                                                                               

                                                                                                                                                                  

 (a) ∫ e¹⁰ˣ d𝓍

Textbook Question

Area functions for constant functions Consider the following functions ƒ and real numbers a (see figure).

(a) Find and graph the area function A(𝓍) = ∫ₐˣ ƒ(t) dt for ƒ.

ƒ(t) = 5 , a = 0

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(a) If ƒ is a constant function on the interval [a,b], then the right and left Riemann sums give the exact value of ∫ₐᵇ ƒ(𝓍) d𝓍, for any positive integer n.