Properties of integrals Consider two functions ƒ and g on [1,6] such that ∫₁⁶ƒ(𝓍) d𝓍 = 10 and ∫₁⁶g(𝓍) d𝓍 = 5, ∫₄⁶ƒ(𝓍) d𝓍 = 5 , and ∫₁⁴g(𝓍) d𝓍 = 2. Evaluate the following integrals.

(f) ∫₄¹ 2f(𝓍) d𝓍

검증된 단계별 안내

Step 1: Recognize that the integral ∫₄¹ 2f(𝓍) d𝓍 involves a constant multiplier. Use the property of integrals that states ∫ₐᵇ c·ƒ(𝓍) d𝓍 = c·∫ₐᵇ ƒ(𝓍) d𝓍, where c is a constant.

Step 2: Rewrite the integral as 2·∫₄¹ ƒ(𝓍) d𝓍 using the property mentioned in Step 1.

Step 3: Notice that the limits of integration are reversed (from 4 to 1 instead of 1 to 4). Use the property of integrals that states ∫ₐᵇ ƒ(𝓍) d𝓍 = -∫ₐᵇ ƒ(𝓍) d𝓍 when the limits are swapped.

Step 4: Apply the property from Step 3 to rewrite the integral as -2·∫₁⁴ ƒ(𝓍) d𝓍.

Step 5: Use the given information that ∫₁⁶ ƒ(𝓍) d𝓍 = 10 and ∫₄⁶ ƒ(𝓍) d𝓍 = 5 to deduce that ∫₁⁴ ƒ(𝓍) d𝓍 = ∫₁⁶ ƒ(𝓍) d𝓍 - ∫₄⁶ ƒ(𝓍) d𝓍. Substitute this value into the expression from Step 4 to complete the setup for evaluation.

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주요 개념

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Properties of Definite Integrals

Definite integrals have several key properties, including linearity, which states that the integral of a sum of functions is the sum of their integrals, and the ability to reverse the limits of integration, which introduces a negative sign. Understanding these properties is essential for manipulating and evaluating integrals effectively.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, stating that if a function is continuous on [a, b], then the integral of its derivative over that interval equals the change in the function's values. This theorem is crucial for evaluating definite integrals and understanding the relationship between a function and its antiderivative.

Integration by Substitution

Integration by substitution is a technique used to simplify the process of evaluating integrals by changing the variable of integration. This method often involves identifying a part of the integrand that can be substituted with a new variable, making the integral easier to solve. Mastery of this technique is important for tackling more complex integrals.

Properties of integrals Consider two functions ƒ and g on [1,6] such that ∫₁⁶ƒ(𝓍) d𝓍 = 10 and ∫₁⁶g(𝓍) d𝓍 = 5, ∫₄⁶ƒ(𝓍) d𝓍 = 5 , and ∫₁⁴g(𝓍) d𝓍 = 2. Evaluate the following integrals.(f) ∫₄¹ 2f(𝓍) d𝓍