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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2:15 minutes

Problem 5.2.58a

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
(a) ∫₀^π/2 (2 sin θ ― cos θ) dθ

Verified step by step guidance

Step 1: Recognize that the given integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 can be used to evaluate the new integral by leveraging the linearity property of integrals.

Step 2: Rewrite the new integral ∫₀^π/2 (2 sin θ ― cos θ) dθ as ∫₀^π/2 (―(cos θ ― 2 sin θ)) dθ by factoring out a negative sign.

Step 3: Use the property of integrals that states ∫ₐᵇ c·f(x) dx = c·∫ₐᵇ f(x) dx, where c is a constant. Apply this to factor out the negative sign, resulting in ―∫₀^π/2 (cos θ ― 2 sin θ) dθ.

Step 4: Substitute the value of the given integral I = ∫₀^π/2 (cos θ ― 2 sin θ) dθ = ―1 into the expression. This gives ―(―1).

Step 5: Simplify the expression to find the value of the new integral. The result will be the negation of the given integral's value.

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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 and symmetry, allow us to manipulate and evaluate integrals more easily. For instance, the linearity property states that the integral of a sum is the sum of the integrals, and constants can be factored out. Understanding these properties is crucial for simplifying complex integrals and relating them to known values.

Definite Integrals

A definite integral represents the signed area under a curve between two limits. In this case, the integral is evaluated from 0 to π/2, which means we are interested in the behavior of the function within this interval. Knowing how to compute definite integrals and interpret their results is essential for solving the given problem.

Substitution in Integrals

Substitution is a technique used to simplify integrals by changing the variable of integration. This method can help transform a complex integral into a more manageable form. In the context of the given problem, recognizing how to relate the integrals through substitution can lead to an easier evaluation based on the known value of the first integral I.

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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.

(d) ∫₄⁶ (g(𝓍) ― f(𝓍) d𝓍

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