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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 5.2.59
Chapter 5, Problem 5.2.59

Definite integrals from graphs The figure shows the areas of regions bounded by the graph of Ζ’ and the 𝓍-axis. Evaluate the following integrals.




βˆ«β‚€α΅ƒ Ζ’(𝓍) d𝓍

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Identify the regions bounded by the graph of the function Ζ’(𝓍) and the 𝓍-axis between 𝓍 = 0 and 𝓍 = a. Note whether these regions lie above or below the 𝓍-axis, as this affects the sign of the integral.
Recall that the definite integral βˆ«β‚€α΅ƒ Ζ’(𝓍) d𝓍 represents the net area between the curve and the 𝓍-axis from 0 to a. Areas above the axis contribute positively, while areas below contribute negatively.
Determine the numerical values of the areas shown in the figure for each region between 0 and a. Assign positive values to areas above the axis and negative values to areas below the axis.
Sum all these signed areas to find the value of the definite integral βˆ«β‚€α΅ƒ Ζ’(𝓍) d𝓍. This sum represents the net area under the curve from 0 to a.
Write the final expression for the integral as the sum of these areas, ensuring to include the correct signs based on the position relative to the 𝓍-axis.

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Key Concepts

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

Definite Integral as Net Area

A definite integral from a to b of a function f(x) represents the net area between the graph of f and the x-axis over [a, b]. Areas above the x-axis contribute positively, while areas below contribute negatively, resulting in the integral's value.
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Interpreting Graphs for Integration

When evaluating integrals from graphs, it is essential to identify the regions bounded by the curve and the x-axis, noting their shapes and whether they lie above or below the axis. This helps in calculating areas accurately, often by summing or subtracting given region areas.
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Properties of Definite Integrals

Definite integrals have properties such as additivity over intervals and sign changes when limits are reversed. Understanding these properties allows breaking complex integrals into simpler parts and correctly combining areas to find the total integral value.
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Related Practice
Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 βˆ« sec 4w tan 4w dw

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Textbook Question

Symmetry in integrals Use symmetry to evaluate the following integrals.

βˆ«Β²β‚‹β‚‚ [(xΒ³ ― 4x) / (xΒ² + 1)] dx 

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Textbook Question

Average value of the derivative Suppose Ζ’ ' is a continuous function for all real numbers. Show that the average value of the derivative on an interval [a, b] is ƒ⁻' = (Ζ’(b) ―ƒ(a))/ (b―a) . Interpret this result in terms of secant lines.

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Textbook Question

General results Evaluate the following integrals in which the function Ζ’ is unspecified. Note that ƒ⁽ᡖ⁾ is the pth derivative of Ζ’ and Ζ’α΅– is the pth power of Ζ’. Assume Ζ’ and its derivatives are continuous for all real numbers. 

∫ (5 Ζ’Β³ (𝓍) + 7Ζ’Β² (𝓍) + Ζ’ (𝓍 )) Ζ’'(𝓍) d𝓍

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Textbook Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  

                                                                                                                                                                    

 βˆ« (sin⁡ 𝓍 + 3 sinΒ³ 𝓍― sin 𝓍) cos 𝓍 d𝓍

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Textbook Question

Use the given substitution to evaluate the following indefinite integrals. Check your answer by differentiating.                                                                                              

                                                                                                                                                                                        

 βˆ« (6𝓍 + 1) √(3𝓍² + 𝓍) d𝓍 , u = 3𝓍² + 𝓍

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