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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.2.57b

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals. 
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(b) ∫₁⁰ (2𝓍―𝓍³) d𝓍

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Step 1: Recognize that the integral given in part (b) is the reverse of the integral provided in part (I). Specifically, ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 is the same as ∫₀¹ (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, ∫ₐᵇ f(𝓍) d𝓍 = -∫ᵇₐ f(𝓍) d𝓍.
Step 3: Apply this property to the given integral. Since ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4, swapping the limits gives ∫₁⁰ (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, ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 = -∫₁⁰ (𝓍³ ― 2𝓍) d𝓍.
Step 5: Substitute the value of ∫₁⁰ (𝓍³ ― 2𝓍) d𝓍 from Step 3 into the equation from Step 4. This gives ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 = -3/4.

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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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