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Lesson 5-1

Lesson 5-1. Area Underneath the Curve. Ice Breaker. Sketch the graph and use geometry to find the area:. y. 3. -1. x. 5. -1. 4 - x² 0 ≤ x ≤ 2 f(x) = x - 2 2 < x ≤ 5. Area = ¼ (4 π ) + ½ (9) = 7.642. Objectives.

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Lesson 5-1

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  1. Lesson 5-1 Area Underneath the Curve

  2. Ice Breaker • Sketch the graph and use geometry to find the area: y 3 -1 x 5 -1 4 - x² 0 ≤ x ≤ 2 f(x) = x - 2 2 < x ≤ 5 Area = ¼ (4π) + ½ (9) = 7.642

  3. Objectives • Find the area underneath a curve using limits • Find the distance traveled by an object (like a car)

  4. Vocabulary • Area problem – find the area under the curve (and the x-axis) between two endpoints • Area – is the limit (as n approaches infinity) of the sum of n rectangles • Distance problem – find the area under the velocity curve (and the x-axis) between two endpoints

  5. a b Area Under the Curve How do we find the area of the shaded region?

  6. Rectangles from Midpoints 1 2 5 3 4 N = 5 a b The area in yellow is the underestimations and the area in blue are the overestimations of the area. Midpoints seem to give better estimates than either inscribed or circumscribed rectangles.

  7. Trapezoidal Estimates 5 1 2 3 4 N = 5 a b The area in yellow is the underestimations and the area in blue are the overestimations of the area. Trapezoids also give better estimates than either inscribed or circumscribed rectangles.

  8. Example 2a Estimate the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] with 5 subintervals using inscribed rectangles. xi = a + (i-1)∆x ∆x=(b-a)/n Area of Rectangle = l ∙ w Ri = (F(xi)) ∙ (∆x) y 5 5 R1 = (1+ 0²) ∙ (0.4) = 0.4 R2 = (1+ 0.4²) ∙ (0.4) = 0.464 4 R3 = (1+ 0.8²) ∙ (0.4) = 0.656 3 2 R4 = (1+ 1.2²) ∙ (0.4) = 0.976 x 0 0 2 1 R5 = (1+ 1.6²) ∙ (0.4) = 1.424 ∑Ri= 3.92 True Area = 4.666

  9. Example 2b Estimate the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] with 5 subintervals using circumscribed rectangles. xi = a + (i)∆x ∆x=(b-a)/n Area of Rectangle = l ∙ w 5 Ri = (F(xi)) ∙ (∆x) y 5 4 R1 = (1+ 0.4²) ∙ (0.4) = 0.464 R2 = (1+ 0.8²) ∙ (0.4) = 0.656 3 R3 = (1+ 1.2²) ∙ (0.4) = 0.976 2 1 R4 = (1+ 1.6²) ∙ (0.4) = 1.424 x 0 0 2 R5 = (1+ 2²) ∙ (0.4) = 2.0 ∑Ri= 5.52 True Area = 4.666

  10. Example 2c Estimate the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] with 5 subintervals using midpoint rectangles. xi = a + (i)∆x/2 ∆x=(b-a)/n Area of Rectangle = l ∙ w 5 Ri = (F(xi)) ∙ (∆x) y 5 R1 = (1+ 0.2²) ∙ (0.4) = 0.416 4 R2 = (1+ 0.6²) ∙ (0.4) = 0.544 3 R3 = (1+ 1.0²) ∙ (0.4) = 0.8 2 R4 = (1+ 1.4²) ∙ (0.4) = 1.184 x 0 1 0 2 R5 = (1+ 1.8²) ∙ (0.4) = 1.696 ∑Ri= 4.64 True Area = 4.666

  11. Example 2d Estimate the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] with 5 subintervals using trapezoids. xi = a + (i)∆x ∆x=(b-a)/n Area of Trapezoid = ½ (b1+b2) ∙ w 5 Ti = ½ (F(xi)+F(xi+1)) ∙ (∆x) y 5 4 T1 = ½ (1 + 1.16) ∙ (0.4) = 0.432 T2 = ½ (1.16+ 1.64) ∙ (0.4) = 0.56 3 T3 = ½ (1.64+ 2.44) ∙ (0.4) = 0.816 2 1 T4 = ½ (2.44+ 3.56) ∙ (0.4) = 1.2 x 0 0 2 T5 = ½ (3.46+ 5) ∙ (0.4) = 1.692 ∑Ti= 4.7 True Area = 4.666

  12. Example 2e Use sums to describe the area of the region between the graph of y = x² + 1 and the x-axis from x = 0 to x = 2. Partition [0,2] into n intervals, the width of the intervals will be (2-0)/n = 2/n. Since the function is increasing on this interval, the inscribed heights will be f(xi-1) and the circumscribed heights will be f(xi). (2/n) (1) (2/n) f(0) (2/n) f(0+2/n) (2/n) (1 + (2/n)²) (2/n) f(0+2(2/n)) (2/n) f(0+1(2/n)) (2/n) (1 + (4/n)²) (2/n) f(0+2(2/n)) (2/n) f(0+3(2/n)) (2/n) (1 + (6/n)²) (2/n) f(0+3(2/n)) (2/n) f(0+4(2/n)) (2/n) (1 + (8/n)²) (2/n) f(0+5(2/n)) (2/n) f(0+4(2/n)) (2/n) f(0+(i)(2/n)) (2/n) f(0+(i-1)(2/n)) (2/n) (1 + ((2i-2)/n)²)

  13. Example 3 Find the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] using limits. Lim ∑Ai = Lim ∑f(xi)∆x n→∞ n→∞ ∆x = (2-0)/n = 2/n f(xi) = 1 + (2i/n)² = 1 + 4i²/n² y 5 Ai = 2/n (1 + 4i²/n²) Lim ∑ (2/n + 8i²/n³) n→∞ x 0 0 2 Lim (2/n³) ∑ (n² + 4i²) = Lim (2/n³) (n³ + 4(n³/3 + n²/2 + n/6)) = Lim (2 + 8/3 + 4/n + 8/6n²) = 4.67 n→∞ n→∞ n→∞

  14. Example 4 Estimate the distance traveled (area bounded by the velocity v(x) = 100 – 32t and the x-axis) on the interval [1,3] with 4 subintervals.. xi = a + (i-1)∆x circumscribed xi = a + i∆x inscribed ∆x=(b-a)/n v 100 Area of Rectangle = l ∙ w Ri = (F(xi)) ∙ (∆x) ∆x = 2/4 = 1/2 F(xi) = 100 – 32(xi) t 0 0 5 Circumscribed ½ (68+52+36+20) Inscribed ½ (52+36+20+4) Circumscribed ∑Ri= 93 Inscribed ∑Ri= 56

  15. Summary & Homework • Summary: • Inscribed Rectangles under estimate the area under the curve • Circumscribed Rectangles over estimate the area under the curve • Distance traveled is the area under the velocity curve • In order to find the area underneath a curve we must take a limit of the sum of rectangles as the number of rectangles approaches infinity • Homework: • pg 378 – 380: 2, 5, 12, 13, 15

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