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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 5.2.57b

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. 
I = βˆ«β‚€ΒΉ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(b) βˆ«β‚β° (2𝓍―𝓍³) d𝓍

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1
Step 1: Recognize that the integral given in part (b) is the reverse of the integral provided in part (I). Specifically, βˆ«β‚β° (2𝓍 ― 𝓍³) d𝓍 is the same as βˆ«β‚€ΒΉ (𝓍³ ― 2𝓍) d𝓍, but with the limits of integration swapped.
Step 2: Use the property of integrals that states swapping the limits of integration changes the sign of the integral. Mathematically, βˆ«β‚α΅‡ f(𝓍) d𝓍 = -βˆ«α΅‡β‚ f(𝓍) d𝓍.
Step 3: Apply this property to the given integral. Since βˆ«β‚€ΒΉ (𝓍³ ― 2𝓍) d𝓍 = ―3/4, swapping the limits gives βˆ«β‚β° (𝓍³ ― 2𝓍) d𝓍 = 3/4.
Step 4: Notice that the integrand in part (b) is written as (2𝓍 ― 𝓍³), which is the negative of (𝓍³ ― 2𝓍). Therefore, βˆ«β‚β° (2𝓍 ― 𝓍³) d𝓍 = -βˆ«β‚β° (𝓍³ ― 2𝓍) d𝓍.
Step 5: Substitute the value of βˆ«β‚β° (𝓍³ ― 2𝓍) d𝓍 from Step 3 into the equation from Step 4. This gives βˆ«β‚β° (2𝓍 ― 𝓍³) d𝓍 = -3/4.

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Key Concepts

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

Properties of Integrals

The properties of integrals, such as linearity and the additive property, allow us to manipulate and evaluate integrals more easily. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. Understanding these properties is essential for simplifying complex integrals and relating them to known values.
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Definite Integrals

A definite integral represents the signed area under a curve between two limits. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. Knowing how to evaluate definite integrals is crucial for solving problems that involve finding the total accumulation of quantities over an interval.
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Substitution Method

The substitution method is a technique used to simplify the evaluation of integrals by changing variables. This method involves substituting a part of the integral with a new variable, which can make the integral easier to solve. Mastery of this technique is important for tackling integrals that are not straightforward or that involve complex expressions.
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Related Practice
Textbook Question

{Use of Tech} Functions defined by integrals Consider the function g, which is given in terms of a definite integral with a variable upper limit.


(b) Calculate g'(𝓍)


g(𝓍) = βˆ«β‚€Λ£ sin (Ο€tΒ² ) dt ( a Fresnel integral) 

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Textbook Question

Properties of integrals Use only the fact that βˆ«β‚€β΄ 3𝓍 (4 ―𝓍) d𝓍 = 32, and the definitions and properties of integrals, to evaluate the following integrals, if possible.

(b) βˆ«β‚€β΄ 𝓍(𝓍 ― 4) d(𝓍)

Textbook Question

{Use of Tech} Approximating net area The following functions are positive and negative on the given interval.


f(x) = sin 2x on [0,3Ο€/4]


(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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Textbook Question

The following functions are positive and negative on the given interval.

Ζ’(𝓍) = xe⁻ˣ on [-1,1]

(b) Approximate the net area bounded by the graph of f and the x-axis on the interval using a left, right, and midpoint Riemann sum with n = 4.

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Textbook Question

{Use of Tech} Riemann sums for larger values of n Complete the following steps for the given function f and interval.


Ζ’(𝓍) = xΒ² ― 1 on [2,5] ; n = 75

(b) Based on the approximations found in part (a), estimate the area of the region bounded by the graph of f and the x-axis on the interval.

Textbook Question

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

(b) A left Riemann sum always overestimates the area of a region bounded by a positive increasing function and the x-axis on an interval [a,b].

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