1. Lesson 6-1b
Area Between Curves

2. Ice Breaker Homework Check Reading questions: none
(Setup only) Find the area between y = 9 – x², y = ½ x and the lines x=-1 and x =2
Reading questions: none

3. Objectives Find the area between two curves
Using either x (vertical rectangles) or
y (horizontal rectangles) as the variable of integration

4. Vocabulary
Area under curve – definite integral of the curve evaluated between two endpoints
Area between curves – definite integral of the height (usually the difference in the curves) evaluated between two point (can be points of intersection or just two arbitrary points)

5. ∫ f(x) dx = Area = ∑ rectangles (f(x)dx)
Area under the Curve a b
f(x) dx
∫ f(x) dx = Area = ∑ rectangles (f(x)dx)
x = a
x = b

6. Example 4 ∫ f(x) dx = ∫ f(x) dx + ∫ (0 – f(x)) dx1 dx
f(x)
f(x) = x3 – 3x2 – x + 3
∫ f(x) dx = ∫ f(x) dx + ∫ (0 – f(x)) dx
= ∫ (x3 – 3x2 – x + 3) dx – ∫ (x3 – 3x2 – x + 3) dx
= ¼x4 – x3 - ½x2 + 3x | – ¼x4 – x3 - ½x2 + 3x |
= [(¼ ½ + 3) – (¼ ½ - 3)]
- [( ) – (¼ ½ + 3)]
= [7/4 – (-9/4)] – [(0) – (7/4)]
= /4 = 23/4 = 5 ¾
x = -1
x = 2
x = 1 6

7. Example 5 Using horizontal rectangles, dy Area = ∫ (g(y) - f(y)) dy-1 dx dy
g(y) = x = 3 – y2
g(x) = +/- √3-x
f(x) = y = x – 1
f(y) = x = y + 1 3
y = -2
y = 1
Area = ∫ (g(y) - f(y)) dy
= ∫ ((3 - y²) – (y + 1)) dy
= ∫ (2 – y - y²) dy
= (2y – ½y² - ⅓y³) |
= (2 – ½ - ⅓) – ( -4 – 2 – (-8/3))
= (7/6) – (-10/3) = 27/6 = 9/2
Using horizontal rectangles, dy
Area = ∫ (f(x)-g(x)) dx + 2∫ (-g(x)) dx
x = 2
x = 3
x = -1
Using vertical rectangles, dx
Note: the reason for the two integrals is that at x = 2 the top curve changes from f(x) to g(x) 7

8. Example 6 Using horizontal rectangles, dy Area = ∫ (f(y) - g(y)) dy
g(y) = x = y2
f(y) = x = 3y + 4 4 -1
Area = ∫ (f(y) - g(y)) dy
= ∫ ((3y + 4) – (y²)) dy
= ∫ (4 + 3y - y²) dy
= (4y + (3/2)y² - ⅓y³) |
= ( – 64/3) – ( /2 – (-1/3))
= (56/3) – (-13/6) = 125/6
y = -1
y = 4
Using horizontal rectangles, dy
Step 1: Find intersection points
Step 2: Rough Graph
Step 3: Determine vertical (top – bottom) dx or horizontal (right – left) dy
Step 4: Solve 8

9. ∫ (h(y) – j(y)) dy (horizontal rectangles)
In General
right curve – left curve
top curve – bottom curve
j(y)
g(x) d dx
h(y) – j(y)
f(x)
f(x) – g(x) dy
h(y) c a b
∫ (h(y) – j(y)) dy (horizontal rectangles)
y = c
y = d
∫ (f(x) – g(x)) dx (vertical rectangles)
x = a
x = b

10. Example 7 ∫ (y – ½(y + 1)) dy ∫ (x –x²) dx (vertical rectangles)
right curve – left curve
top curve – bottom curve
f(x) =x
h(y) – j(y) dx 1
g(x) = x² dy
f(x) – g(x) 1
∫ (y – ½(y + 1)) dy
(horizontal rectangles)
y = 0
y = 1
∫ (x –x²) dx (vertical rectangles)
x = 0
x = 1

11. Summary & Homework Summary: Homework: pg 442 - 442: 2, 12, 15, 17
Area between curves is still a height times a width
Width is always dx (vertical) or dy (horizontal)
Height is the difference between the curves or between a curve and a line
Homework:
pg : 2, 12, 15, 17