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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2:42 minutes

Problem 5.2.57a

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
(a) ∫₀¹ (4𝓍―2𝓍³) d𝓍

Verified step by step guidance

Step 1: Recognize that the given integral in part (a), ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍, can be split into two separate integrals using the linearity property of integrals: ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍 = ∫₀¹ 4𝓍 d𝓍 ― ∫₀¹ 2𝓍³ d𝓍.

Step 2: Factor out constants from each integral. For the first term, ∫₀¹ 4𝓍 d𝓍 becomes 4∫₀¹ 𝓍 d𝓍. For the second term, ∫₀¹ 2𝓍³ d𝓍 becomes 2∫₀¹ 𝓍³ d𝓍.

Step 3: Use the given value of I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4 to extract the value of ∫₀¹ 𝓍³ d𝓍. Rewrite I as ∫₀¹ 𝓍³ d𝓍 ― ∫₀¹ 2𝓍 d𝓍 = ―3/4. This equation can be solved to find ∫₀¹ 𝓍³ d𝓍.

Step 4: Compute ∫₀¹ 𝓍 d𝓍 using the power rule for integration. The integral of 𝓍 with respect to 𝓍 is (𝓍²)/2. Evaluate this from 0 to 1 to find the value of ∫₀¹ 𝓍 d𝓍.

Step 5: Substitute the values of ∫₀¹ 𝓍³ d𝓍 and ∫₀¹ 𝓍 d𝓍 into the expression for ∫₀¹ (4𝓍 ― 2𝓍³) d𝓍 = 4∫₀¹ 𝓍 d𝓍 ― 2∫₀¹ 𝓍³ d𝓍 to compute the final result.

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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 ability to split integrals, are fundamental in calculus. Linearity allows us to factor constants out of integrals and combine integrals of the same limits. This means that if we have an integral of a sum, we can separate it into the sum of integrals, which simplifies calculations.

Definite Integrals

Definite integrals represent the signed area under a curve between two points on the x-axis. The notation ∫ₐᵇ f(x) dx indicates the integral of the function f(x) from a to b. The value of a definite integral can be interpreted as the accumulation of quantities, which is essential for evaluating integrals over specific intervals.

Integration Techniques

Various techniques exist for evaluating integrals, including substitution and integration by parts. In this context, recognizing patterns in the integrand can help simplify the integral. For example, if an integral can be expressed in terms of known integrals, such as the one provided (I), it can be evaluated more easily by leveraging previously calculated values.

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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.

(a) ∫₁⁴ 3f(𝓍) d𝓍

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