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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1:37 minutes

Problem 5.2.57b

Textbook Question

Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(b) ∫₁⁰ (2𝓍―𝓍³) d𝓍

Verified step by step guidance

Step 1: Recognize that the integral given in part (b) is the reverse of the integral provided in part (I). Specifically, ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 is the same as ∫₀¹ (𝓍³ ― 2𝓍) d𝓍, but with the limits of integration swapped.

Step 2: Use the property of integrals that states swapping the limits of integration changes the sign of the integral. Mathematically, ∫ₐᵇ f(𝓍) d𝓍 = -∫ᵇₐ f(𝓍) d𝓍.

Step 3: Apply this property to the given integral. Since ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4, swapping the limits gives ∫₁⁰ (𝓍³ ― 2𝓍) d𝓍 = 3/4.

Step 4: Notice that the integrand in part (b) is written as (2𝓍 ― 𝓍³), which is the negative of (𝓍³ ― 2𝓍). Therefore, ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 = -∫₁⁰ (𝓍³ ― 2𝓍) d𝓍.

Step 5: Substitute the value of ∫₁⁰ (𝓍³ ― 2𝓍) d𝓍 from Step 3 into the equation from Step 4. This gives ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 = -3/4.

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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 the additive property, allow us to manipulate and evaluate integrals more easily. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. Understanding these properties is essential 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. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. Knowing how to evaluate definite integrals is crucial for solving problems that involve finding the total accumulation of quantities over an interval.

Substitution Method

The substitution method is a technique used to simplify the evaluation of integrals by changing variables. This method involves substituting a part of the integral with a new variable, which can make the integral easier to solve. Mastery of this technique is important for tackling integrals that are not straightforward or that involve complex expressions.

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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.

(b) ∫₁⁶ (f(𝓍) ― g(𝓍)) d𝓍

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