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Part1. Below, we see two graphs with the area under each curve shaded in blue. These shaded areas are equal as a result of making a substitution, u =x + 1 in the integral This graph of depicts the area under the curve y = x + 1 from x = 0 to a = 4 which is the value of the definite integral This graph of depicts the area under the curve y = 2u from u = 1 to u = 3 which is the value of the definite integral Part 2. In general, when making a change of variables in a definite integral we must change the limits of integration appropriately. Suppose that we let u = x + 1 so that 2du = dx. Then, Using the formula for area of a trapezoid, the two shaded region can be shown to have equal area. This implies that the two definite integrals must be equal. That is, [₁² ( 1² x + 1) dx = ²° ²0 น when = 0 and u = when x = 4 ( 7² +1) d² Part 3. Show that the two shaded areas above are equal by evaluating the following definite integrals. (²x -0 x + 1) dx = [2u du -0-0 2u du ¹ / ² ( ² ² + ¹ ) dz 2u du

Part1. Below, we see two graphs with the area under each curve shaded in blue. These shaded areas are equal as a result of making a substitution, u =x + 1 in the integral This graph of depicts the area under the curve y = x + 1 from x = 0 to a = 4 which is the value of the definite integral This graph of depicts the area under the curve y = 2u from u = 1 to u = 3 which is the value of the definite integral Part 2. In general, when making a change of variables in a definite integral we must change the limits of integration appropriately. Suppose that we let u = x + 1 so that 2du = dx. Then, Using the formula for area of a trapezoid, the two shaded region can be shown to have equal area. This implies that the two definite integrals must be equal. That is, [₁² ( 1² x + 1) dx = ²° ²0 น when = 0 and u = when x = 4 ( 7² +1) d² Part 3. Show that the two shaded areas above are equal by evaluating the following definite integrals. (²x -0 x + 1) dx = [2u du -0-0 2u du ¹ / ² ( ² ² + ¹ ) dz 2u du

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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Part1. Below, we see two graphs with the area under each curve shaded in blue. These shaded areas are equal as a result of making a substitution, u = This graph of depicts the area under the curve y = x + 1 from x = 0 to x = 4 which is the value of the definite integral This graph of depicts the area under the curve y = 2u from u = 1 to u = 3 which is the value of the definite integral Part 2. In general, when making a change of variables in a definite integral we must change the limits of integration appropriately. Suppose that we let u = +1 so that 2du = dx. Then, u Using the formula for area of a trapezoid, the two shaded region can be shown to have equal area. This implies that the two definite integrals must be equal. That is, [*^(1/² + 1) dx = [ 2u dus when = 0 and u [²² Part 3. Show that the two shaded areas above are equal by evaluating the following definite integrals. √² (²+1) dr == 2u du when = 4 = [(x + 1) de 11-0 f 2u du + 1 in the integral ¹ / ² ( 1/2 ² + ¹ ) d ²
Transcribed Image Text:Part1. Below, we see two graphs with the area under each curve shaded in blue. These shaded areas are equal as a result of making a substitution, u = This graph of depicts the area under the curve y = x + 1 from x = 0 to x = 4 which is the value of the definite integral This graph of depicts the area under the curve y = 2u from u = 1 to u = 3 which is the value of the definite integral Part 2. In general, when making a change of variables in a definite integral we must change the limits of integration appropriately. Suppose that we let u = +1 so that 2du = dx. Then, u Using the formula for area of a trapezoid, the two shaded region can be shown to have equal area. This implies that the two definite integrals must be equal. That is, [*^(1/² + 1) dx = [ 2u dus when = 0 and u [²² Part 3. Show that the two shaded areas above are equal by evaluating the following definite integrals. √² (²+1) dr == 2u du when = 4 = [(x + 1) de 11-0 f 2u du + 1 in the integral ¹ / ² ( 1/2 ² + ¹ ) d ²
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