Lesson 6-1b Area Between Curves

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

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

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)

∫ 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

Example 4 ∫ f(x) dx = ∫ f(x) dx + ∫ (0 – f(x)) dx 1 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 - 1 - ½ + 3) – (¼ + 1 - ½ - 3)] - [(4 - 8 - 2 + 6) – (¼ - 1 - ½ + 3)] = [7/4 – (-9/4)] – [(0) – (7/4)] = 4 + 7/4 = 23/4 = 5 ¾ x = -1 x = 2 x = 1 6

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

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) | = (16 + 24 – 64/3) – ( -4 + 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

∫ (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

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

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 442 - 442: 2, 12, 15, 17