Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

8. Definite Integrals

Introduction to Definite Integrals

Calculus

8. Definite Integrals

Introduction to Definite Integrals

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1:59 minutes

Problem 5.2.58b

Textbook Question

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1
(b) ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ

Verified step by step guidance

Step 1: Recognize that the given integral (b) ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ can be expressed in terms of the original integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1.

Step 2: Factor out the constant multiplier from the integrand in (b). Using the property of integrals, ∫ₐᵇ k * f(x) dx = k * ∫ₐᵇ f(x) dx, rewrite the integral as ∫₀^π/2 (4 cos θ ― 8 sin θ) dθ = 4 * ∫₀^π/2 (cos θ ― 2 sin θ) dθ.

Step 3: Substitute the value of the original integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 into the expression derived in Step 2.

Step 4: Multiply the constant factor (4) by the value of the original integral (―1) to simplify the expression.

Step 5: The result of the multiplication gives the value of the integral (b).

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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, allow us to manipulate integrals in useful ways. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. This is crucial for evaluating integrals efficiently, especially when they can be expressed in terms of known integrals.

Definite Integrals

Definite integrals represent the signed area under a curve between two limits. In this case, the limits are from 0 to π/2. Understanding how to compute definite integrals and their properties is essential for evaluating integrals like the one given in the question, as it provides the numerical value of the area.

Substitution in Integrals

Substitution is a technique used to simplify the evaluation of integrals by changing the variable of integration. This method can transform a complex integral into a simpler form, making it easier to compute. Recognizing when and how to apply substitution is key to solving integrals effectively.

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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

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