Textbook Question
Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.
(d) ∫₆³ (ƒ(𝓍) + 2g(𝓍)) d𝓍
56
views
8. Definite Integrals
Introduction to Definite Integrals
2:54 minutesProblem 5.2.55f
Textbook Question
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
Verified step by step guidance1
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.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
Recommended video:
05:43
Definition of the Definite Integral
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.
Recommended video:
06:11Fundamental Theorem of Calculus Part 1
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.
Recommended video:
04:27Substitution With an Extra Variable
Related Videos
Related Practice