y = x - 2 x +1 + x 4 y = 5 - x For the given figure, notice that the shaded region is in the interval [0, 4]); however, neither graph lies above the other over this entire interval. It appears that the lines intersect at x = 2, but this should be confirmed algebraically. To find the x-coordinates of the intersection points, set the equations y = x2 - 2x + 1 and y = 5 - x² equal to each other and solve for x as follows. x2 - 2x + 1 = 5 - x2 2x2 - 2x – 4 = 0 collect like terms x2 - x - 2 = o divide by 2 (x + 1) = 0 factor x + 1 = 0 or x - 2 = 0 set each factor equal to zero x = -1 or x = solve for x Since x = -1 is not in the interval [0, 4], this solution is discarded. The solution x = the x-coordinate of the intersection point of the two graphs in the interval [0, 4]. gives us We can now apply the area formula above if we split the interval [0, 4] into two intervals, one from [0, 2] and the other from

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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For the given figure, notice that the shaded region is in the interval [0, 4];however, neither graph lies above the other over this entire interval. It appears that the lines intersect at x = 2, but this should be confirmed algebraically. To find the x-coordinates of the intersection points, set the equations y = x2 − 2x + 1 and y = 5 − x2 equal to each other and solve for x as follows.

 

see picture answer whats on  yellow 

**Example Problem: Calculating the Area Between Two Curves**

**Graph Description:**

The graph given displays two parabolic curves intersecting. The first curve, labeled \( y = x^2 - 2x + 1 \), opens upwards, representing a standard quadratic function. The second curve, labeled \( y = 5 - x^2 \), opens downwards. The region bounded by these two curves, between \( x = 0 \) and \( x = 4 \), is shaded in blue.

**Explanation:**

For the given figure, notice that the shaded region is in the interval [0, 4]; however, neither graph lies above the other over this entire interval. It appears that the lines intersect at \( x = 2 \), but this should be confirmed algebraically. To find the x-coordinates of the intersection points, set the equations \( y = x^2 - 2x + 1 \) and \( y = 5 - x^2 \) equal to each other and solve for \( x \) as follows.

\[ x^2 - 2x + 1 = 5 - x^2 \]

\[ 2x^2 - 2x - 4 = 0 \quad \text{(collect like terms)} \]

\[ x^2 - x - 2 = 0 \quad \text{(divide by 2)} \]

\[ (x + 1)(x - 2) = 0 \quad \text{(factor)} \]

\[ x + 1 = 0 \quad \text{or} \quad x - 2 = 0 \quad \text{(set each factor equal to zero)} \]

\[ x = -1 \quad \text{or} \quad x = 2 \quad \text{(solve for x)} \]

Since \( x = -1 \) is not in the interval [0, 4], this solution is discarded. The solution \( x = 2 \) gives us the x-coordinate of the intersection point of the two graphs in the interval [0, 4].

We can now apply the area formula above if we split the interval [0, 4] into two intervals, one from [0, 2] and the other from \([2, 4]\).

For further detailed steps and computations, continue exploring additional resources on integration techniques
Transcribed Image Text:**Example Problem: Calculating the Area Between Two Curves** **Graph Description:** The graph given displays two parabolic curves intersecting. The first curve, labeled \( y = x^2 - 2x + 1 \), opens upwards, representing a standard quadratic function. The second curve, labeled \( y = 5 - x^2 \), opens downwards. The region bounded by these two curves, between \( x = 0 \) and \( x = 4 \), is shaded in blue. **Explanation:** For the given figure, notice that the shaded region is in the interval [0, 4]; however, neither graph lies above the other over this entire interval. It appears that the lines intersect at \( x = 2 \), but this should be confirmed algebraically. To find the x-coordinates of the intersection points, set the equations \( y = x^2 - 2x + 1 \) and \( y = 5 - x^2 \) equal to each other and solve for \( x \) as follows. \[ x^2 - 2x + 1 = 5 - x^2 \] \[ 2x^2 - 2x - 4 = 0 \quad \text{(collect like terms)} \] \[ x^2 - x - 2 = 0 \quad \text{(divide by 2)} \] \[ (x + 1)(x - 2) = 0 \quad \text{(factor)} \] \[ x + 1 = 0 \quad \text{or} \quad x - 2 = 0 \quad \text{(set each factor equal to zero)} \] \[ x = -1 \quad \text{or} \quad x = 2 \quad \text{(solve for x)} \] Since \( x = -1 \) is not in the interval [0, 4], this solution is discarded. The solution \( x = 2 \) gives us the x-coordinate of the intersection point of the two graphs in the interval [0, 4]. We can now apply the area formula above if we split the interval [0, 4] into two intervals, one from [0, 2] and the other from \([2, 4]\). For further detailed steps and computations, continue exploring additional resources on integration techniques
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