6.1 Area Between Curves

Notes

1. Calculating area of a region as the difference of two areas
   1. sketch the graphs of y = x² + 3, y = x/2 + 1, x = 0, and x = 4. Lightly shade the area enclosed by these graphs.
      This is a [-0.5,5] x [-1,25] window.
      
      1. Demonstrate the "Shade" operation on the TI-89 (F5 Menu - A) .
         
   2. A Riemann approximation
      1. Approximate the area between the graph of y = x² + 3 and the x-axis on [0, 4] using a Riemann sum with n = 4 and left-hand endpoints. Be sure to draw the rectangles! Call this approximation A_top.
         (A_top = 1(3 + 4 + 7 + 12) = 26
         
      2. Review the Riemann sum function on the TI-89.
         
      3. Approximate the area between the graph of y = x/2 + 1 and the x-axis on [0, 4] using a Riemann sum with n = 4 and left-hand endpoints. Be sure to draw the rectangles! Call this approximation A_bottom.
         A_bottom = 1(1 + 3/2 + 3 + 5/2) = 7
         
      4. A = A_top - A_bottom.
         
         A = 26 - 7 = 19
         
   3. Using a definite integral
      1. Write a definite integral that expresses the area between the graph of y = x² + 3 and the x-axis on [0, 4]. Emphasize how the definite integral expresses the sum of rectangles. Evaluate this integral. Call this area A_top.
         A_top = 100/3 = 33.333
         
      2. Write a definite integral that expresses the area between the graph of y = x/2 + 1 and the x-axis on [0, 4] and left-hand endpoints. Emphasize how the definite integral expresses the sum of rectangles. Evaluate this integral. Call this approximation A_bottom.
         A_bottom = 8
         
      3. A = A_top - A_bottom
         
         A=100/3 - 8 = 76/3 = 25.333
         
   4. Limitation: The definite integral only represents the area between the function and the x-axis where the function is non-negative.
      
2. Calculating area of a region as a single area
   1. Regraph the functions in part A.
      
   2. A Riemann approximation
      1. Approximate the area between the graph of y = x² + 3 and the graph of y = x/2 + 1 on [0, 4] using a Riemann sum with n = 4. Be sure to draw the rectangles!
         A = 1( 2 + 5/2 + 5 + 19/2) = 19
         
   3. Using a definite integral
      1. Write a definite integral that expresses the area between the graph of y = x² + 3 and the graph of y = x/2 + 1 on [0, 4]. Emphasize how the definite integral expresses the sum of rectangles. Evaluate this integral.
         A = 76/3 = 25.333
3. One function has some negative values
   1. Find the area enclosed by the graphs of y = x² + 3, y = 1- x, x = 0, and x = 4.
      1. Sketch the graphs and locate the enclosed area.
         
      2. Draw some representative rectangles.
         
         1. Finding the height of the rectangles
            
         2. Finding the base of the rectangles
            
      3. Construct a definite integral
         
      4. Evaluate
         
         
4. Both functions have negative values
   1. Find the area enclosed by the graphs of y = -1 - x², y = -x - 4, x = -1, and x = 1.
      1. Sketch the graphs and locate the enclosed area.
         Window is [-2, 2] x [-6, 1]
         
      2. Construct the definite integral and evaluate.
         
5. Finding points of intersection
   1. Find the area enclosed by the graphs of y = x² and y = x.
      1. Sketch the graphs and locate the enclosed area.
         Window is [-.2, 1.2] x [-.2, 1.2]
         
      2. Where do the graphs intersect? Solve x² = x for x. Factor x² - x = 0.
         
      3. Construct the definite integral and evaluate.
         
         
   2. Find the area enclosed by the graphs of y = -1 - x², y = -x - 4.
      1. Sketch the graphs and locate the enclosed area.
         Window is [-2, 3] x [-8,1]
         
      2. Where do the graphs intersect? -1 - x² = -x - 4
         
      3. Construct the definite integral and evaluate.
         
6. Multi-part Areas
   1. Find the area enclosed by the graphs of y = sin x and y = x/2 between x = 0 and x = .
      1. Sketch the graphs and locate the enclosed areas.
         Window is [0, 3.5] x [-0.2, 2.2]
         
      2. Where do the graphs intersect? sin x = x
         x = 1.895
         
      3. Construct the definite integrals and evaluate.
         
7. Areas bounded by relations
   1. Find the area enclosed by the graphs of x = y² - 4y and x = 2y - y². (This is problem #4, p. 308 in Stewart)
      1. Method 1 - the direct approach
         1. Sketch the graphs and locate the enclosed area. Here, your calculator won't help! Plot some points and use your knowledge of the graphs of parabolas.
            
         2. Where do the graphs intersect? Where y² - 4y = 2y - y².
            y = 0 or y = 3, so the points of intersection are (0, 0) and (-3, 3)
            
         3. Construct the definite integral and evaluate.
            
      2. Method 2 - using the inverse
         1. Find the inverse of each relation by switching all y's to x's and switching all x's to y's.
            y = x² - 4x and y = 2x - x².
            
         2. Proceed as before
            Window is [-1, 4] x [-8, 5]
            
            
   2. Find the area enclosed by the graphs of y = (x + 1)^1/5 and x = 3y - y².
      1. Even though only one equation is a relation, you can still find the inverse of each.
         x = (y + 1)^1/5 , so y = x⁵ - 1, and y = 3x - x ².
         
      2. Proceed as before:
         Window is [-1.5, 1.5] x [-7, 2.5]
         x⁵ - 1 = 3x - x ² gives x = -1.383, -0.303, 1.261
         

last update January 1, 2007 by JL Stanbrough