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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.3.51

Definite integrals Evaluate the following integrals using the Fundamental Theorem of Calculus


βˆ«β‚β΄ (𝓍 ― 2)/βˆšπ“ d𝓍

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Step 1: Recognize that the integral βˆ«β‚β΄ (𝓍 ― 2)/βˆšπ“ d𝓍 can be simplified by breaking the integrand into separate terms. Rewrite the integrand as (𝓍/βˆšπ“) - (2/βˆšπ“). This simplifies to βˆšπ“ - 2/βˆšπ“.
Step 2: Split the integral into two separate integrals: βˆ«β‚β΄ βˆšπ“ d𝓍 - βˆ«β‚β΄ (2/βˆšπ“) d𝓍. This allows us to evaluate each term individually.
Step 3: For the first integral βˆ«β‚β΄ βˆšπ“ d𝓍, rewrite βˆšπ“ as 𝓍^(1/2). Use the power rule for integration: βˆ«π“βΏ d𝓍 = (𝓍^(n+1))/(n+1) + C, where n β‰  -1. Apply this rule to find the antiderivative of 𝓍^(1/2).
Step 4: For the second integral βˆ«β‚β΄ (2/βˆšπ“) d𝓍, rewrite 2/βˆšπ“ as 2𝓍^(-1/2). Again, use the power rule for integration to find the antiderivative of 𝓍^(-1/2).
Step 5: Apply the Fundamental Theorem of Calculus to both antiderivatives. Evaluate the antiderivatives at the upper limit (𝓍 = 4) and subtract the value at the lower limit (𝓍 = 1). Combine the results to find the value of the definite integral.

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Key Concepts

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

Definite Integrals

Definite integrals represent the signed area under a curve between two specified limits. They are calculated using the integral symbol with lower and upper bounds, indicating the interval over which the function is evaluated. The result of a definite integral is a numerical value that quantifies this area, which can be interpreted in various contexts, such as physics and economics.
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Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with integration, providing a method to evaluate definite integrals. It states that if a function is continuous on an interval, then the integral of its derivative over that interval equals the difference in the values of the original function at the endpoints. This theorem allows for the computation of definite integrals by finding antiderivatives.
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Antiderivatives

An antiderivative of a function is another function whose derivative is the original function. Finding an antiderivative is a key step in evaluating definite integrals using the Fundamental Theorem of Calculus. For example, if F(x) is an antiderivative of f(x), then the definite integral from a to b can be computed as F(b) - F(a), providing a straightforward way to find the area under the curve.
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Related Practice
Textbook Question

Use symmetry to explain why.

βˆ«β΄β‚‹β‚„ (5𝓍⁴ + 3𝓍³ + 2𝓍² + 𝓍 + 1) d𝓍 = 2 βˆ«β‚€β΄ (5𝓍⁴ + 2𝓍² + 𝓍 + 1) d𝓍 .

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Textbook Question

Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.                                                                                                                         

                                                                                                                                                                              

 βˆ«β‚‚/β‚β‚…βˆšβ‚ƒβ‚Ž^Β²/⁡ d𝓍/ x√(25𝓍²― 1)

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Textbook Question

Area functions from graphs The graph of Ζ’ is given in the figure. A(𝓍) = βˆ«β‚€Λ£ Ζ’(t) dt and evaluate A(2), A(5), A(8), and A(12).


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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« 𝓍 csc 𝓍² cot 𝓍² d𝓍

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Textbook Question

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βˆ«β΄β‚‹β‚† √(24 ― 2𝓍 ― 𝓍²) d𝓍

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Textbook Question

{Use of Tech} Areas of regions Find the area of the region 𝑅 bounded by the graph of Ζ’ and the 𝓍-axis on the given interval. Graph Ζ’ and show the region 𝑅.                                              

                                                                                                                                                                                    

 Ζ’(𝓍) = 𝓍² (𝓍 ― 2) on [ ―1 , 3]

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