chapter 14 11

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MATH 1325 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 1 MATH 1325 14.11 Area between Curves

14.11 Area between Curves :Chapter 14.11 © Rita Marie O’Brien 2005 Page 2 14.11 Area between Curves Find the area of the region bounded by the graphs of the given conditions. Be sure to find any needed points of intersection. Sketch the functions first. Find the points of intersection Integrate

14.11 Area between Curves :Chapter 14.11 © Rita Marie O’Brien 2005 Page 3 14.11 Area between Curves If the variable is x, to find the area between to curves use the following procedure.

14.11 Area between Curves :Chapter 14.11 © Rita Marie O’Brien 2005 Page 4 14.11 Area between Curves If the variable is y, to find the area between to curves use the following procedure.

14.11 Problem 2 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 5 14.11 Problem 2 y = 2x ( a line ) y = x2 ( a parabola ) We need to find the x – values of the points of intersection. To find the x – value of the intersection points, set the two functions equal to each other and solve for x. The green shaded part is the area we are to find.

Find the x – values of the points of intersection. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 6 Find the x – values of the points of intersection. Continue to next slide to check your answer. Problem 2 continued

14.11 Problem 2 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 7 14.11 Problem 2 continued 2x = x2 -2x -2x 0 = x2 - 2x Factor: 0 = x(x-2) 0 = x, 0 = x – 2 x = 0, x = 2 We know this is true from the picture provided in the book. (see next slide) Plug in the x –value to either one of the functions to find the y-value. Using y = 2x y = 2(0) = 0 Point of intersection is (0,0) y = 2(2) = 4 Point of intersection is (2,4)

14.11 Problem 2 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 8 14.11 Problem 2 continued graph

Set up the integral. Do Not Evaluate!Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 9 Set up the integral. Do Not Evaluate!Continue to next slide to check your answer. Problem 2 continued

14.11 Problem 2 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 10 14.11 Problem 2 continued Formula:

14.11 Problem 4 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 11 14.11 Problem 4 y = x ( a line ) y = x(x-2)2 ( a cubic ) We need to find the points of intersection. x = x(x-2)2 -x -x 0 = x(x-2)2 – x Factor: 0 = x [ (x-2)2 – 1] 0 = x, 0 = (x-2)2 – 1 0 = (x-2)2 – 1 0 = (x-2)2 – 1 Add 1 to both sides: 1 = (x-2)2 Take the square root of both sides: +/-1 = (x-2) + 1 = x - 2, -1 = x - 2 3 = x, 1 = x The x-values for the points of intersection are 0,1, and 3.

14.11 Problem 4 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 12 14.11 Problem 4 continued Graph: In the yellow shaded areas y = x(x-2)2 is on top. In the green shaded areas y = x is on top We calculate the area between the points of intersection. There are two areas.

14.11 Problem 4 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 13 14.11 Problem 4 continued Graph: In the yellow shaded area y = x(x-2)2 is on top. In the green shaded area y = x is on top We need two integrals.

Set up the integrals. Do Not Evaluate!Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 14 Set up the integrals. Do Not Evaluate!Continue to next slide to check your answer. Problem 4 continued

14.11 Problem 4 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 15 14.11 Problem 4 continued Formula:

14.11 Problem 6 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 16 14.11 Problem 6 y = 4 y = - 2x – 8 y = 2x We need to find the points of intersection. There are three. Where y = 4 and y = 2x meet, Where y = 4 and y = - 2x - 8 meet Where y = 2x and y = - 2x - 8 meet Case 1: 4 = 2x 2 = x Case 2: 4 = - 2x - 8 12 = - 2x - 6 = x Case 3: 2x = - 2x - 8 4x = - 8 x = - 2

14.11 Problem 6 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 17 14.11 Problem 6 continued Graph:

14.11 Problem 6 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 18 14.11 Problem 6 continued Graph: The top function is always y = 4 Which function is on the bottom depends on where you are looking. Between – 6 and – 2, y = - 2x – 8 is the bottom Between – 2 and 2, y = 2x is the bottom We would need to break the area into two parts.

14.11 Problem 6 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 19 14.11 Problem 6 continued Graph: If we choose to work with the variable y then we are concerned with the right and left functions. If we choose this method we do not need to break up the area into two parts. This would be more efficient.

14.11 Problem 6 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 20 14.11 Problem 6 y = 4 y = - 2x – 8 y = 2x We need to find the y value of the points of intersection. X = - 6,- 2, and 2 For the intersection of y = 4 with y = 2x, it is obvious that y = 4. For the intersection of y = 4 with y = - 2x - 8, it is obvious that y = 4. X = - 2 this happens when y = 2x and y = - 2x - 8 intersect. Use y = 2x to find the y-value of the point of intersection y = 2( -2) = - 4 Point of intersection is (- 2, - 4)

14.11 Problem 6 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 21 14.11 Problem 6 To use y as the variable we must solve for x. y = - 2x – 8 y = 2x y + 8 = - 2x Divide by – 2:

14.11 Problem 6 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 22 14.11 Problem 6 continued Integral: Do Not Evaluate!

14.11 Problem 12 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 23 14.11 Problem 12 y = x2 +1 ( a parabola ) y = x + 3 ( a line ) We need to find the x – values of the points of intersection. To find the intersection points, set the two functions equal to each other and solve for x.

Find the x – values of the points of intersection. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 24 Find the x – values of the points of intersection. Continue to next slide to check your answer. Problem 12 continued

14.11 Problem 12 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 25 14.11 Problem 12 continued y = x2 +1 ( a parabola ) y = x + 3 ( a line ) We need to find the x- values of the points of intersection. x2 + 1 = x + 3 x2 – x – 2 = 0 (x + 1) (x – 2) = 0 x = - 1, x = 2

Graph. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 26 Graph. Continue to next slide to check your answer. Problem 12 continued

14.11 Problem 12 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 27 14.11 Problem 12 continued y = x2 +1 ( a parabola ) y = x + 3 ( a line ) The green shaded part is the area we are to find.

Set up the integral. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 28 Set up the integral. Continue to next slide to check your answer. Problem 12 continued

14.11 Problem 12 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 29 14.11 Problem 12 continued y = x2 +1 ( a parabola ) y = x + 3 ( a line ) The green shaded part is the area we are to find.

Evaluate the integral. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 30 Evaluate the integral. Continue to next slide to check your answer. Problem 12 continued

14.11 Problem 12 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 31 14.11 Problem 12 continued Evaluate:

14.11 Problem 12 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 32 14.11 Problem 12 continued continued.

14.11 Problem 12 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 33 14.11 Problem 12 continued continued.

14.11 Problem like 26 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 34 14.11 Problem like 26 y2 = 2 - x ( a parabola on its side ) y = x + 4 ( a line ) x = y2 looks like: We need to find the area using the right – left, which means working with the variable y, so solve the equations for x.

Solve the equations for x. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 35 Solve the equations for x. Continue to next slide to check your answer. Problem like 26 continued

14.11 Problem like 26 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 36 14.11 Problem like 26 y2 = 2 - x ( a parabola on its side ) y = x + 4 ( a line ) Solve for x: y2 = 2 – x y = x + 4 y2 + x = 2 y – 4 = x x = 2 - y2

14.11 Problem like 26 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 37 14.11 Problem like 26 x = 2 - y2 y - 4 = x We need to find the y – values of the points of intersection. To find the y values of the intersection points, set the two functions equal to each other and solve for y.

Find the y – values of the points of intersection. Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 38 Find the y – values of the points of intersection. Continue to next slide to check your answer. Problem like 26 continued

14.11 Problem like 26 :Chapter 14.11 © Rita Marie O’Brien 2005 Page 39 14.11 Problem like 26 x = 2 - y2 y - 4 = x 2 - y2 = y – 4 0 = y2 + y -6 0 = (y+3)(y-2) 0 = y + 3, 0 = y – 2 y = - 3, y = 2

14.11 Problem like 26 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 40 14.11 Problem like 26 continued x = 2 - y2 y - 4 = x The green shaded part is the area we are to find.

Set up the integral Continue to next slide to check your answer. :Chapter 14.11 © Rita Marie O’Brien 2005 Page 41 Set up the integral Continue to next slide to check your answer. Problem like 26 continued

14.11 Problem like 26 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 42 14.11 Problem like 26 continued x = 2 - y2 y - 4 = x The green shaded part is the area we are to find.

14.11 Problem like 26 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 43 14.11 Problem like 26 continued Evaluate:

14.11 Problem like 26 continued :Chapter 14.11 © Rita Marie O’Brien 2005 Page 44 14.11 Problem like 26 continued Evaluate: