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For the integral be rewritten as av 1² 20x³ sin (5x + 6) dx, we can choose an appropriate function u so the integral can ["sin(u)du.. What are the new limits on this integral? b a

For the integral be rewritten as av 1² 20x³ sin (5x + 6) dx, we can choose an appropriate function u so the integral can ["sin(u)du.. What are the new limits on this integral? b a

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Substitution in Definite Integrals** Given the integral \(\int_{1}^{2} 20x^3 \sin(5x^4 + 6) \, dx\), we can choose an appropriate substitution function \(u\) so that the integral can be rewritten in terms of \(u\). After substituting, the integral will become: \[ \int_{a}^{b} \sin(u) \, du \] We need to determine the new limits \(a\) and \(b\) based on our substitution. Boxes are provided to input the new limits after substitution: - \(a = \) [Input Box] - \(b = \) [Input Box] To find \(a\) and \(b\), substitute the original integration limits into the function: 1. \( u = 5x^4 + 6 \) 2. When \( x = 1 \): \( u = 5(1)^4 + 6 = 5 + 6 = 11 \) 3. When \( x = 2 \): \( u = 5(2)^4 + 6 = 5(16) + 6 = 80 + 6 = 86 \) So, the new integration limits are: - \( a = 11 \) - \( b = 86 \) Therefore: \[ \int_{1}^{2} 20x^3 \sin(5x^4 + 6) \, dx = \int_{11}^{86} \sin(u) \, du \]
Transcribed Image Text:**Substitution in Definite Integrals** Given the integral \(\int_{1}^{2} 20x^3 \sin(5x^4 + 6) \, dx\), we can choose an appropriate substitution function \(u\) so that the integral can be rewritten in terms of \(u\). After substituting, the integral will become: \[ \int_{a}^{b} \sin(u) \, du \] We need to determine the new limits \(a\) and \(b\) based on our substitution. Boxes are provided to input the new limits after substitution: - \(a = \) [Input Box] - \(b = \) [Input Box] To find \(a\) and \(b\), substitute the original integration limits into the function: 1. \( u = 5x^4 + 6 \) 2. When \( x = 1 \): \( u = 5(1)^4 + 6 = 5 + 6 = 11 \) 3. When \( x = 2 \): \( u = 5(2)^4 + 6 = 5(16) + 6 = 80 + 6 = 86 \) So, the new integration limits are: - \( a = 11 \) - \( b = 86 \) Therefore: \[ \int_{1}^{2} 20x^3 \sin(5x^4 + 6) \, dx = \int_{11}^{86} \sin(u) \, du \]
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