lesson 5-1
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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. - PowerPoint PPT PresentationTRANSCRIPT
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Lesson 5-1
Area Underneath the Curve
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Ice Breaker• Sketch the graph and use geometry to find the
area:y
x5
3
-1
-1
4 - x² 0 ≤ x ≤ 2f(x) = x - 2 2 < x ≤ 5
Area = ¼ (4π) + ½ (9) = 7.642
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Objectives
• Find the area underneath a curve using limits
• Find the distance traveled by an object (like a car)
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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
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How do we find the area of the shaded region?
a b
Area Under the Curve
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Rectangles from Midpoints
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.
N = 5
a b
1 23 4
5
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Trapezoidal Estimates
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.
N = 51 23 4
5
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Example 2aEstimate 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.y
x2
5
00
1 2 34
5
Area of Rectangle = l ∙ w
R1 = (1+ 0²) ∙ (0.4) = 0.4
Ri = (F(xi)) ∙ (∆x)
R2 = (1+ 0.4²) ∙ (0.4) = 0.464
R3 = (1+ 0.8²) ∙ (0.4) = 0.656
R4 = (1+ 1.2²) ∙ (0.4) = 0.976
R5 = (1+ 1.6²) ∙ (0.4) = 1.424
∑Ri = 3.92
xi = a + (i-1)∆x ∆x=(b-a)/n
True Area = 4.666
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Example 2bEstimate 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.y
x2
5
00
1
5
4
32
Area of Rectangle = l ∙ w
R1 = (1+ 0.4²) ∙ (0.4) = 0.464
Ri = (F(xi)) ∙ (∆x)
R2 = (1+ 0.8²) ∙ (0.4) = 0.656
R3 = (1+ 1.2²) ∙ (0.4) = 0.976
R4 = (1+ 1.6²) ∙ (0.4) = 1.424
R5 = (1+ 2²) ∙ (0.4) = 2.0
∑Ri = 5.52
xi = a + (i)∆x ∆x=(b-a)/n
True Area = 4.666
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Example 2cEstimate 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.y
x2
5
00
1
5
43
2
Area of Rectangle = l ∙ w
R1 = (1+ 0.2²) ∙ (0.4) = 0.416
Ri = (F(xi)) ∙ (∆x)
R2 = (1+ 0.6²) ∙ (0.4) = 0.544
R3 = (1+ 1.0²) ∙ (0.4) = 0.8
R4 = (1+ 1.4²) ∙ (0.4) = 1.184
R5 = (1+ 1.8²) ∙ (0.4) = 1.696
∑Ri = 4.64
xi = a + (i)∆x/2 ∆x=(b-a)/n
True Area = 4.666
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Example 2dEstimate the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] with 5 subintervals using trapezoids.y
x2
5
00
1 23
4
5
Area of Trapezoid = ½ (b1+b2) ∙ w
T1 = ½ (1 + 1.16) ∙ (0.4) = 0.432
Ti = ½ (F(xi)+F(xi+1)) ∙ (∆x)
T2 = ½ (1.16+ 1.64) ∙ (0.4) = 0.56
T3 = ½ (1.64+ 2.44) ∙ (0.4) = 0.816
T4 = ½ (2.44+ 3.56) ∙ (0.4) = 1.2
T5 = ½ (3.46+ 5) ∙ (0.4) = 1.692
∑Ti = 4.7
xi = a + (i)∆x ∆x=(b-a)/n
True Area = 4.666
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Example 2eUse 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).
Rectangle Inscribed Area Circumscribed Area1
2
3
4
5
i
(2/n) f(0) (2/n) f(0+2/n)
(2/n) f(0+1(2/n)) (2/n) f(0+2(2/n))(2/n) f(0+2(2/n)) (2/n) f(0+3(2/n))(2/n) f(0+3(2/n)) (2/n) f(0+4(2/n))(2/n) f(0+4(2/n)) (2/n) f(0+5(2/n))(2/n) f(0+(i-1)(2/n)) (2/n) f(0+(i)(2/n))
(2/n) (1)
(2/n) (1 + (2/n)²)(2/n) (1 + (4/n)²)(2/n) (1 + (6/n)²)(2/n) (1 + (8/n)²)(2/n) (1 + ((2i-2)/n)²)
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Example 3Find the area bounded by the function f(x) = x² + 1 and the x-axis on the interval [0,2] using limits.y
x2
5
00
∆x = (2-0)/n = 2/nf(xi) = 1 + (2i/n)² = 1 + 4i²/n²
Ai = 2/n (1 + 4i²/n²)
Lim ∑Ai = Lim ∑f(xi)∆xn→∞ n→∞
Lim ∑ (2/n + 8i²/n³)n→∞
Lim (2/n³) ∑ (n² + 4i²)n→∞
= Lim (2/n³) (n³ + 4(n³/3 + n²/2 + n/6))n→∞
= Lim (2 + 8/3 + 4/n + 8/6n²) = 4.67n→∞
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Example 4Estimate the distance traveled (area bounded by the velocity v(x) = 100 – 32t and the x-axis) on the interval [1,3] with 4 subintervals..v
t5
100
00
Area of Rectangle = l ∙ w
Ri = (F(xi)) ∙ (∆x)
Circumscribed ∑Ri = 93Inscribed ∑Ri = 56
xi = a + (i-1)∆x circumscribed xi = a + i∆x inscribed
∆x=(b-a)/n
∆x = 2/4 = 1/2
F(xi) = 100 – 32(xi)
Circumscribed ½ (68+52+36+20)Inscribed ½ (52+36+20+4)
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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