Textbook Question
Evaluate ∫₀² 3𝓍² d𝓍 and ∫₋₂² 3𝓍² d𝓍.
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8. Definite Integrals
Introduction to Definite Integrals
2:39 minutesProblem 5.2.55a
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.
(a) ∫₁⁴ 3f(𝓍) d𝓍
Verified step by step guidance1
Step 1: Recall the property of integrals that allows constants to be factored out. Specifically, for any constant c and function ƒ(𝓍), ∫ₐᵇ cƒ(𝓍) d𝓍 = c∫ₐᵇ ƒ(𝓍) d𝓍.
Step 2: Apply this property to the given integral ∫₁⁴ 3ƒ(𝓍) d𝓍. Factor out the constant 3, so the integral becomes 3∫₁⁴ ƒ(𝓍) d𝓍.
Step 3: Notice that the integral ∫₁⁴ ƒ(𝓍) d𝓍 is not directly provided in the problem. However, you can calculate it using the additive property of integrals: ∫₁⁶ ƒ(𝓍) d𝓍 = ∫₁⁴ ƒ(𝓍) d𝓍 + ∫₄⁶ ƒ(𝓍) d𝓍.
Step 4: Substitute the known values into the equation. From the problem, ∫₁⁶ ƒ(𝓍) d𝓍 = 10 and ∫₄⁶ ƒ(𝓍) d𝓍 = 5. Solve for ∫₁⁴ ƒ(𝓍) d𝓍 by subtracting: ∫₁⁴ ƒ(𝓍) d𝓍 = 10 - 5.
Step 5: Replace ∫₁⁴ ƒ(𝓍) d𝓍 in the expression 3∫₁⁴ ƒ(𝓍) d𝓍 with the calculated value from Step 4. The final integral is now expressed as 3 multiplied by the result of ∫₁⁴ ƒ(𝓍) d𝓍.
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Key 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. Additionally, the integral of a constant multiplied by a function can be factored out, allowing for simplification in calculations. Understanding these properties is essential for evaluating integrals efficiently.
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